Chapter I
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Chapter I
Linear Equations
1
I Linear Equations
I.1 Solving Linear Equations
Prerequisites and Learning Goals
From your work in previous courses, you should be able to
• Write a system of linear equations using matrix notation.
• Use Gaussian elimination to bring a system of linear equations into upper triangular
form and reduced row echelon form (rref).
• Determine whether a system of equations has a unique solution, infinitely many solutions or no solutions, and compute all solutions if they exist; express the set of all
solutions in parametric form.
• Compute the inverse of a matrix when it exists, use the inverse to solve a system of
equations, describe for what systems of equations this is possible.
• Find the transpose of a matrix.
• Interpret a matrix as a linear transformation acting on vectors.
After completing this section, you should be able to
• Calculate the standard Euclidean norm, the 1-norm and the infinity norm of a vector.
• Calculate the Hilbert-Schmidt norm of a matrix.
• Define the matrix norm of a matrix; describe the connection between the matrix norm
and how a matrix stretches the length of vectors; compute the matrix norm of a diagonal
matrix.
• Define the condition number of a matrix and its relation to the matrix norm; use
the condition number to estimate relative errors in the solution to a system of linear
equations.
• Explain why a small condition number is desirable in practical computations.
• Use MATLAB/Octave to enter matrices and vectors, make larger matrices from smaller
blocks, multiply matrices, compute the inverse and transpose, extract elements, rows,
columns and submatrices, use rref() to find the reduced row echelon form for a matrix,
solve linear equations using A\b, use rand() to generate random matrices, use tic()
and toc() to time operations, compute norms and condition numbers.
• Use MATLAB/Octave to test conjectures about norms, condition numbers, etc.
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I.1 Solving Linear Equations
I.1.1 Review: Systems of linear equations
The first part of the course is about systems of linear equations. You will have studied such
systems in a previous course, and should remember how to find solutions (when they exist)
using Gaussian elimination.
Many practical problems can be solved by turning them into a system of linear equations. In
this chapter we will study a few examples: the problem of finding a function that interpolates
a collection of given points, and the approximate solutions of differential equations. In
practical problems, the question of existence of solutions, although important, is not the
end of the story. It turns out that some systems of equations, even though they may have
a unique solution, are very sensitive to changes in the coefficients. This makes them very
difficult to solve reliably. We will see some examples of such ill-conditioned systems, and
learn how to recognize them using the condition number of a matrix.
Recall that a system of linear equations, like this system of 2 equations in 3 unknowns
x1 +2x2 +x3 = 0
x1 −5x2 +x3 = 1
can be written as a matrix equation
x1
1 2 1
0
x2 =
.
1 −5 1
1
x3
A general system of m linear equations in n unknowns can be written as
Ax = b
where A is an given m × n (m rows, n columns) matrix, b is a given m-component vector,
and x is the n-component vector of unknowns.
A system of linear equations may have no solutions, a unique solutions, or infinitely many
solutions. This is easy to see when there is only a single variable x, so that the equation has
the form
ax = b
where a and b are given numbers. The solution is easy to find if a 6= 0: x = b/a. If a = 0
then the equation reads 0x = b. In this case, the equation either has no solutions (when
b 6= 0) or infinitely many (when b = 0), since in this case every x is a solution.
To solve a general system Ax = b, form the augmented matrix [A|b] and use Gaussian
elimination to reduce the matrix to reduced row echelon form. This reduced matrix (which
represents a system of linear equations that has exactly the same solutions as the original
system) can be used to decide whether solutions exist, and to find them. If you don’t
remember this procedure, you should review it.
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I Linear Equations
In the example above, the augmented matrix is
1 2 1 0
.
1 −5 1 1
The reduced row echelon form is
1 0 1 2/7
,
0 1 0 −1/7
which leads to a family of solutions (one for each value of the parameter s)
2/7
−1
x = −1/7 + s 0 .
0
1
I.1.2 Solving a non-singular system of n equations in n unknowns
Let’s start with a system of equations where the number of equations is the same as the
number of unknowns. Such a system can be written as a matrix equation
Ax = b,
where A is a square matrix, b is a given vector, and x is the vector of unknowns we are
trying to find. When A is non-singular (invertible) there is a unique solution. It is given by
x = A−1 b, where A−1 is the inverse matrix of A. Of course, computing A−1 is not the most
efficient way to solve a system of equations.
For our first introduction to MATLAB/Octave, let’s consider an example:
1 1
1
3
A = 1 1 −1 b = 1 .
1
1 −1 1
First, we define the matrix A and the vector b in MATLAB/Octave. Here is the input (after
the prompt symbol >) and the output (without a prompt symbol).
>A=[1 1 1;1 1 -1;1 -1 1]
A =
1
1
1
1
1
-1
1
-1
1
>b=[3;1;1]
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I.1 Solving Linear Equations
b =
3
1
1
Notice that the entries on the same row are separated by spaces (or commas) while rows
are separated by semicolons. In MATLAB/Octave, column vectors are n by 1 matrices and
row vectors are 1 by n matrices. The semicolons in the definition of b make it a column
vector. In MATLAB/Octave, X’ denotes the transpose of X. Thus we get the same result if
we define b as
>b=[3 1 1]’
b =
3
1
1
The solution can be found by computing the inverse of A and multiplying
>x = A^(-1)*b
x =
1
1
1
However if A is a large matrix we don’t want to actually calculate the inverse. The syntax
for solving a system of equations efficiently is
>x = A\b
x =
1
1
1
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I Linear Equations
If you try this with a singular matrix A, MATLAB/Octave will complain and print an
warning message. If you see the warning, the answer is not reliable! You can always check
to see that x really is a solution by computing Ax.
>A*x
ans =
3
1
1
As expected, the result is b.
By the way, you can check to see how much faster A\b is than A^(-1)*b by using the
functions tic() and toc(). The function tic() starts the clock, and toc() stops the clock
and prints the elapsed time. To try this out, let’s make A and b really big with random
entries.
A=rand(1000,1000);
b=rand(1000,1);
Here we are using the MATLAB/Octave command rand(m,n) that generates an m × n
matrix with random entries chosen between 0 and 1. Each time rand is used it generates
new numbers.
Notice the semicolon ; at the end of the inputs. This suppresses the output. Without the
semicolon, MATLAB/Octave would start writing the 1,000,000 random entries of A to our
screen! Now we are ready to time our calculations.
tic();A^(-1)*b;toc();
Elapsed time is 44 seconds.
tic();A\b;toc();
Elapsed time is 13.55 seconds.
So we see that A\b quite a bit faster.
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I.1 Solving Linear Equations
I.1.3 Reduced row echelon form
How can we solve Ax = b when A is singular, or not a square matrix (that is, the number
of equations is different from the number of unknowns)? In your previous linear algebra
course you learned how to use elementary row operations to transform the original system
of equations to an upper triangular system. The upper triangular system obtained this way
has exactly the same solutions as the original system. However, it is much easier to solve.
In practice, the row operations are performed on the augmented matrix [A|b].
If efficiency is not an issue, then addition row operations can be used to bring the system
into reduced row echelon form. In the this form, the pivot columns have a 1 in the pivot
position and zeros elsewhere. For example, if A is a square non-singular matrix then the
reduced row echelon form of [A|b] is [I|x], where I is the identity matrix and x is the
solution.
In MATLAB/Octave you can compute the reduced row echelon form in one step using the
function rref(). For the system we considered above we do this as follows. First define A
and b as before. This time I’ll suppress the output.
>A=[1 1 1;1 1 -1;1 -1 1];
>b=[3 1 1]’;
In MATLAB/Octave, the square brackets [ ... ] can be used to construct larger matrices
from smaller building blocks, provided the sizes match correctly. So we can define the
augmented matrix C as
>C=[A b]
C =
1
1
1
1
1
-1
1
-1
1
3
1
1
Now we compute the reduced row echelon form.
>rref(C)
ans =
1
0
0
0
1
0
0
-0
1
1
1
1
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I Linear Equations
The solution appears on the right.
Now let’s try to solve Ax = b with
1 2 3
A = 4 5 6
7 8 9
1
b = 1 .
1
This time the matrix A is singular and doesn’t have an inverse. Recall that the determinant
of a singular matrix is zero, so we can check by computing it.
>A=[1 2 3; 4 5 6; 7 8 9];
>det(A)
ans = 0
However we can still try to solve the equation Ax = b using Gaussian elimination.
>b=[1 1 1]’;
>rref([A b])
ans =
1.00000
0.00000
0.00000
0.00000
1.00000
0.00000
-1.00000
2.00000
0.00000
-1.00000
1.00000
0.00000
Letting x3 = s be a parameter, and proceeding as you learned in previous courses, we arrive
at the general solution
−1
1
x = 1 + s −2 .
0
1
On the other hand, if
1 2 3
A = 4 5 6
7 8 9
then
>rref([1 2 3 1;4 5 6 1;7 8 9 0])
ans =
1.00000
0.00000
0.00000
0.00000
1.00000
0.00000
-1.00000
2.00000
0.00000
tells us that there is no solution.
8
0.00000
0.00000
1.00000
1
b = 1 ,
0
I.1 Solving Linear Equations
I.1.4 Gaussian elimination steps using MATLAB/Octave
If C is a matrix in MATLAB/Octave, then C(1,2) is the entry in the 1st row and 2nd column.
The whole first row can be extracted using C(1,:) while C(:,2) yields the second column.
Finally we can pick out the submatrix of C consisting of rows 1-2 and columns 2-4 with the
notation C(1:2,2:4).
Let’s illustrate this by performing a few steps of Gaussian elimination on the augmented
matrix from our first example. Start with
C=[1 1 1 3; 1 1 -1 1; 1 -1 1 1];
The first step in Gaussian elimination is to subtract the first row from the second.
>C(2,:)=C(2,:)-C(1,:)
C =
1
0
1
1
0
-1
1
-2
1
3
-2
1
Next, we subtract the first row from the third.
>C(3,:)=C(3,:)-C(1,:)
C =
1
0
0
1
0
-2
1
-2
0
3
-2
-2
To bring the system into upper triangular form, we need to swap the second and third rows.
Here is the MATLAB/Octave code.
>temp=C(3,:);C(3,:)=C(2,:);C(2,:)=temp
C =
1
0
0
1
-2
0
1
0
-2
3
-2
-2
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I Linear Equations
I.1.5 Norms for a vector
Norms are a way of measuring the size of a vector. They are important when we study how
vectors change, or want to know how close one vector is to another. A vector may have many
components and it might happen that some are big and some are small. A norm is a way of
capturing information about the size of a vector in a single number. There is more than one
way to define a norm.
In your previous linear algebra course, you probably have encountered the most common
norm, called the Euclidean norm (or the 2-norm). The word norm without qualification
usually refers to this norm. What is the Euclidean norm of the vector
−4
a=
?
3
When you draw the vector as an arrow on the plane, this norm is the Euclidean distance
between the tip and the tail. This leads to the formula
p
kak = (−4)2 + 32 = 5.
This is the answer that MATLAB/Octave gives too:
> a=[-4 3]
a =
-4
3
> norm(a)
ans = 5
The formula is easily generalized to n dimensions. If x = [x1 , x2 , . . . , xn ]T then
p
kxk = |x1 |2 + |x2 |2 + · · · + |xn |2 .
The absolute value signs in this formula, which might seem superfluous, are put in to make
the formula correct when the components are complex numbers. So, for example
√
i p
√
= |i|2 + |1|2 = 1 + 1 = 2.
1 Does MATLAB/Octave give this answer too?
There are situations where other ways of measuring the norm of a vector are more natural.
Suppose that the tip and tail of the vector a = [−4, 3]T are locations in a city where you can
only walk along the streets and avenues.
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I.1 Solving Linear Equations
3
−4
If you defined the norm to be the shortest distance that you can walk to get from the tail
to the tip, the answer would be
kak1 = | − 4| + |3| = 7.
This norm is called the 1-norm and can be calculated in MATLAB/Octave by adding 1 as
an extra argument in the norm function.
> norm(a,1)
ans = 7
The 1-norm is also easily generalized to n dimensions. If x = [x1 , x2 , . . . , xn ]T then
kxk1 = |x1 | + |x2 | + · · · + |xn |.
Another norm that is often used measures the largest component in absolute value. This
norm is called the infinity norm. For a = [−4, 3]T we have
kak∞ = max{| − 4|, |3|} = 4.
To compute this norm in MATLAB/Octave we use inf as the second argument in the norm
function.
> norm(a,inf)
ans = 4
Here are three properties that the norms we have defined all have in common:
1. For every vector x and every number s, ksxk = |s|kxk.
2. The only vector with norm zero is the zero vector, that is, kxk = 0 if and only if x = 0
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I Linear Equations
3. For all vectors x and y, kx + yk ≤ kxk + kyk. This inequality is called the triangle
inequality. It says that the length of the longest side of a triangle is smaller than the
sum of the lengths of the two shorter sides.
What is the point of introducing many ways of measuring the length of a vector? Sometimes
one of the non-standard norms has natural meaning in the context of a given problem. For
example, when we study stochastic matrices, we will see that multiplication of a vector by a
stochastic matrix preserves the 1-norm of the vector. So in this situation it is natural to use
1-norms. However, in this course we will almost always use the standard Euclidean norm.
If v a vector then kvk (without any subscripts) will always denote the standard Euclidean
norm.
I.1.6 Matrix norms
Just as for vectors, there are many ways to measure the size of a matrix A.
For a start we could think of a matrix as a vector whose entries just happen to be written
in a box, like
1 2
A=
,
0 2
rather than in a row, like
1
2
a=
0 .
2
√
Taking this point of view, we would define the norm of A to be 12 + 22 + 02 + 22 = 3. In
fact, the norm computed in this way is sometimes used for matrices. It is called the HilbertSchmidt norm. For a general matrix A = [ai,j ], the formula for the Hilbert-Schmidt norm
is
sX X
kAkHS =
|ai,j |2 .
i
j
The Hilbert-Schmidt norm does measure the size of matrix in some sense. It has the advantage of being easy to compute from the entries ai,j . But it is not closely tied to the action of
A as a linear transformation.
When A is considered as a linear transformation or operator, acting on vectors, there is
another norm that is more natural to use.
Starting with a vector x the matrix A transforms it to the vector Ax. We want to say that
a matrix is big if increases the size of vectors, in other words, if kAxk is big compared to kxk.
So it is natural to consider the stretching ratio kAxk/kxk. Of course, this ratio depends on
x, since some vectors get stretched more than others by A. Also, the ratio is not defined if
x = 0. But in this case Ax = 0 too, so there is no stretching.
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I.1 Solving Linear Equations
We now define the matrix norm of A to be the largest of these ratios,
kAxk
.
x:kxk6=0 kxk
kAk = max
This norm measures the maximum factor by which A can stretch the length of a vector. It
is sometimes called the operator norm.
Since kAk is defined to be the maximum of a collection of stretching ratios, it must be
bigger than or equal to any particular stretching ratio. In other words, for any non zero
vector x we know kAk ≥ kAxk/kxk, or
kAxk ≤ kAkkxk.
This is how the matrix norm is often used in practice. If we know kxk and the matrix norm
kAk, then we have an upper bound on the norm of Ax.
In fact, the maximum of a collection of numbers is the smallest number that is larger than
or equal to every number in the collection (draw a picture on the number line to see this),
the matrix norm kAk is the smallest number that is bigger than kAxk/kxk for every choice
of non-zero x. Thus kAk is the smallest number C for which
kAxk ≤ Ckxk
for every x.
An equivalent definition for kAk is
kAk = max kAxk.
x:kxk=1
Why do these definitions give the same answer? The reason is that the quantity kAxk/kxk
does not change if we multiply x by a non-zero scalar (convince yourself!). So, when calculating the maximum over all non-zero vectors in the first expression for kAk, all the vectors
pointing in the same direction will give the same value for kAxk/kxk. This means that we
need only pick one vector in any given direction, and might as well choose the unit vector.
For this vector, the denominator is equal to one, so we can ignore it.
Here is another way of saying this. Consider the image of the unit sphere under A. This
is the set of vectors {Ax : kxk = 1} The length of the longest vector in this set is kAk.
The picture
below
is a sketch of the unit sphere (circle) in two dimensions, and its image
1 2
under A =
. This image is an ellipse.
0 2
||A||
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I Linear Equations
The norm of the matrix is the distance from the originq
to the point on the ellipse farthest
√
from the origin. In this case this turns out to be kAk = 9/2 + (1/2) 65.
It’s hard to see how this expression can be obtained from the entries of the matrix. There
is no easy formula. However, if A is a diagonal matrix the norm is easy to compute.
If
To see this, let’s consider a diagonal matrix
3 0 0
A = 0 2 0 .
0 0 1
x1
x = x2
x3
then
3x1
Ax = 2x2
x3
so that
kAxk2 = |3x1 |2 + |2x2 |2 + |x3 |2
= 32 |x1 |2 + 22 |x2 |2 + |x3 |2
≤ 32 |x1 |2 + 32 |x2 |2 + 32 |x3 |2
= 32 kxk2 .
This implies that for any unit vector x
kAxk ≤ 3
and taking the maximum over all unit vectors x yields kAk ≤ 3. On the other hand, the
maximum of kAxk over all unit vectors x is larger than the value of kAxk for any particular
unit vector. In particular, if
1
e1 = 0
0
then
kAk ≥ kAe1 k = 3.
Thus we see that
kAk = 3.
In general, the matrix norm of a diagonal matrix with diagonal entries λ1 , λ2 , · · · , λn is the
largest value of |λk |.
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I.1 Solving Linear Equations
The MATLAB/Octave code for a diagonal matrix with diagonal entries 3, 2 and 1 is
diag([3 2 1]) and the expression for the norm of A is norm(A). So for example
>norm(diag([3 2 1]))
ans =
3
I.1.7 Condition number
Let’s return to the situation where A is a square matrix and we are trying to solve Ax = b.
If A is a matrix arising from a real world application (for example if A contains values
measured in an experiment) then it will almost never happen that A is singular. After all,
a tiny change in any of the entries of A can change a singular matrix to a non-singular one.
What is much more likely to happen is that A is close to being singular. In this case A−1
will still exist, but will have some enormous entries. This means that the solution x = A−1 b
will be very sensitive to the tiniest changes in b so that it might happen that round-off error
in the computer completely destroys the accuracy of the answer.
To check whether a system of linear equations is well-conditioned, we might therefore think
of using kA−1 k as a measure. But this isn’t quite right, since we actually don’t care if kA−1 k
is large, provided it stretches each vector about the same amount. For example, if we simply
multiply each entry of A by 10−6 the size of A−1 will go way up, by a factor of 106 , but our
ability to solve the system accurately is unchanged. The new solution is simply 106 times
the old solution, that is, we have simply shifted the position of the decimal point.
It turns out that for a square matrix A, the ratio of the largest stretching factor to the
smallest stretching factor of A is a good measure of how well conditioned the system of
equation Ax = b is. This ratio is called the condition number and is denoted cond(A).
Let’s first compute an expression for cond(A) in terms of matrix norms. Then we will
explain why it measures the conditioning of a system of equations.
We already know that the largest stretching factor for a matrix A is the matrix norm kAk.
So let’s look at the smallest streching factor. We might as well assume that A is invertible.
Otherwise, there is a non-zero vector that A sends to zero, so that the smallest stretching
factor is 0 and the condition number is infinite.
min
x6=0
kAxk
kAxk
= min
x6=0 kA−1 Axk
kxk
kyk
= min
y6=0 kA−1 yk
1
=
kA−1 yk
max
y6=0
kyk
1
.
=
kA−1 k
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I Linear Equations
Here we used the fact that if x ranges over all non-zero vectors so does y = Ax and that
the minimum of a collection of positive numbers is one divided by the maximum of their
reciprocals. Thus the smallest stretching factor for A is 1/kA−1 k. This leads to the following
formula for the condition number of an invertible matrix:
cond(A) = kAkkA−1 k.
In our applications we will use the condition number as a measure of how well we can solve
the equations that come up accurately.
Now, let us try to see why the condition number of A is a good measure of how well we
can solve the equations Ax = b accurately.
Starting with Ax = b we change the right side to b′ = b + ∆b. The new solution is
x′ = A−1 (b + ∆b) = x + ∆x
where x = A−1 b is the original solution and the change in the solutions is ∆x = A−1 ∆b.
Now the absolute errors k∆bk and k∆xk are not very meaningful, since an absolute error
k∆bk = 100 is not very large if kbk = 1, 000, 000, but is large if kbk = 1. What we really care
about are the relative errors k∆bk/kbk and k∆xk/kxk. Can we bound the relative error
in the solution in terms of the relative error in the equation? The answer is yes. Beginning
with
k∆xkkbk = kA−1 ∆bkkAxk
≤ kA−1 kk∆bkkAkkxk,
we can divide by kbkkxk to obtain
k∆bk
k∆xk
≤ kA−1 kkAk
kxk
kbk
k∆bk
.
= cond(A)
kbk
This equation gives the real meaning of the condition number. If the condition number is
near to 1 then the relative error of the solution is about the same as the relative error in
the equation. However, a large condition number means that a small relative error in the
equation can lead to a large relative error in the solution.
In MATLAB/Octave the condition number is computed using cond(A).
> A=[2 0; 0 0.5];
> cond(A)
ans = 4
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I.1 Solving Linear Equations
I.1.8 Summary of MATLAB/Octave commands used in this section
How to create a row vector
[ ] square brackets are used to construct matrices and vectors. Create a row in the matrix
by entering elements within brackets. Separate each element with a comma or space.
For example, to create a row vector a with three columns (i.e. a 1-by-3 matrix), type
a=[1 1 1] or equivalently a=[1,1,1]
How to create a column vector or a matrix with more than one row
; when the semicolon is used inside square brackets, it terminates rows. For example,
a=[1;1;1] creates a column vector with three rows
B=[1 2 3; 4 5 6] creates a 2 − by − 3 matrix
’ when a matrix (or a vector) is followed by a single quote ’ (or apostrophe) MATLAB
flips rows with columns, that is, it generates the transpose. When the original matrix
is a simple row vector, the apostrophe operator turns the vector into a column vector.
For example,
a=[1 1 1]’ creates a column vector with three rows
B=[1 2 3; 4 5 6]’ creates a 3 − by − 2 matrix where the first row is 1 4
How to use specialized matrix functions
rand(n,m) returns a n-by-m matrix with random numbers between 0 and 1.
How to extract elements or submatrices from a matrix
A(i,j) returns the entry of the matrix A in the i-th row and the j-th column
A(i,:) returns a row vector containing the i-th row of A
A(:,j) returns a column vector containing the j-th column of A
A(i:j,k:m) returns a matrix containing a specific submatrix of the matrix A. Specifically,
it returns all rows between the i-th and the j-th rows of A, and all columns between
the k-th and the m-th columns of A.
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I Linear Equations
How to perform specific operations on a matrix
det(A) returns the determinant of the (square) matrix A
rref(A) returns the reduced row echelon form of the matrix A
norm(V) returns the 2-norm (Euclidean norm) of the vector V
norm(V,1) returns the 1-norm of the vector V
norm(V,inf) returns the infinity norm of the vector V
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I.2 Interpolation
I.2 Interpolation
Prerequisites and Learning Goals
From your work in previous courses, you should be able to
• compute the determinant of a square matrix; apply the basic linearity properties of
the determinant, and explain what its value means about existence and uniqueness of
solutions.
After completing this section, you should be able to
• give a definition of interpolation function and explain the idea of getting a unique
interpolation function by restricting the class of functions under consideration.
• Define the problem of Lagrange interpolation and express it in terms of a system of
equations where the unknowns are the coefficients of a polynomial of given degree; set
up the system in matrix form using the Vandermonde matrix, derive the formula for
the determinant of the Vandermonde matrix; explain why a solution to the Lagrange
interpolation problem always exists.
• Explain why Lagrange interpolation is not a practical method for large numbers of
points.
• Define the mathematical problem of interpolation using splines, compare and contrast
it with Lagrange interpolation.
• Explain how minimizing the bending energy leads to a description of the shape of the
spline as a piecewise polynomial function.
• Express the interpolation problem of cubic splines in terms of a system of equations
where the unknowns are related to the coefficients of the cubic polynomials.
• Given a set of points, use MATLAB/Octave to calculate and plot the interpolating
polynomial in Lagrange interpolation and the piecewise function for cubic splines.
• Use the MATLAB/Octave functions linspace, vander, polyval, zeros and ones.
• Use m files in MATLAB/Octave.
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I Linear Equations
I.2.1 Introduction
Suppose we are given some points (x1 , y1 ), . . . , (xn , yn ) in the plane, where the points xi are
all distinct.
Our task is to find a function f (x) that passes through all these points. In other words, we
require that f (xi ) = yi for i = 1, . . . , n. Such a function is called an interpolating function.
Problems like this arise in practical applications in situations where a function is sampled
at a finite number of points. For example, the function could be the shape of the model we
have made for a car. We take a bunch of measurements (x1 , y1 ), . . . , (xn , yn ) and send them
to the factory. What’s the best way to reproduce the original shape?
Of course, it is impossible to reproduce the original shape with certainty. There are infinitely many functions going through the sampled points.
To make our problem of finding the interpolating function f (x) have a unique solution,
we must require something more of f (x), either that f (x) lies in some restricted class of
functions, or that f (x) is the function that minimizes some measure of “badness”. We will
look at both approaches.
I.2.2 Lagrange interpolation
For Lagrange interpolation, we try to find a polynomial p(x) of lowest possible degree that
passes through our points. Since we have n points, and therefore n equations p(xi ) = yi to
solve, it makes sense that p(x) should be a polynomial of degree n − 1
p(x) = a1 xn−1 + a2 xn−2 + · · · + an−1 x + an
with n unknown coefficients a1 , a2 , . . . , an . (Don’t blame me for the screwy way of numbering
the coefficients. This is the MATLAB/Octave convention.)
20
I.2 Interpolation
The n equations p(xi ) = yi are n linear equations for these unknown coefficients which we
may write as
a
1
n−1
x1n−2 · · · x21 x1 1
x1
a2
y1
xn−1 xn−2 · · · x2 x2 1 .. y2
2
2
2
.
..
= .. .
..
..
.. ..
..
.
.
. .
a
.
. .
n−2
n−1
n−2
2
yn
xn
xn
· · · xn xn 1 an−1
an
Thus we see that the problem of Lagrange interpolation reduces to solving a system of linear
equations. If this system has a unique solution, then there is exactly one polynomial p(x)
of degree n − 1 running through our points. This matrix for this system of equations has a
special form and is called a Vandermonde matrix.
To decide whether the system of equations has a unique solution we need to determine
whether the Vandermonde matrix is invertible or not. One way to do this is to compute the
determinant. It turns out that the determinant of a Vandermonde matrix has a particularly
simple form, but it’s a little tricky to see this. The 2 × 2 case is simple enough:
x1 1
det
= x1 − x2 .
x2 1
To go on to the 3 × 3 case we won’t simply expand the determinant, but recall that the
determinant is unchanged under row (and column) operations of the type ”add a multiple of
one row (column) to another.” Thus if we start with a 3 × 3 Vandermonde determinant, add
−x1 times the second column to the first, and then add −x1 times the third column to the
second, the determinant doesn’t change and we find that
2
x1 x1 1
0
x1 1
0
0
1
det x22 x2 1 = det x22 − x1 x2 x2 1 = det x22 − x1 x2 x2 − x1 1 .
x23 x3 1
x23 − x1 x3 x3 1
x23 − x1 x3 x3 − x1 1
Now we can take advantage of the zeros in the first row, and calculate the determinant by
expanding along the top row. This gives
2
2
x1 x1 1
x2 (x2 − x1 ) x2 − x1
x2 − x1 x2 x2 − x1
2
.
= det
= det
x2 x2 1
det
x3 (x3 − x1 ) x3 − x1
x23 − x1 x3 x3 − x1
x23 x3 1
Now, we recall that the determinant is linear in each row separately. This implies that
x2 (x2 − x1 ) x2 − x1
x2
1
det
= (x2 − x1 ) det
x3 (x3 − x1 ) x3 − x1
x3 (x3 − x1 ) x3 − x1
x2 1
= (x2 − x1 )(x3 − x1 ) det
.
x3 1
But the determinant on the right is a 2 × 2 Vandermonde determinant that we have already
21
I Linear Equations
computed. Thus we end up with the formula
2
x1 x1 1
det x22 x2 1 = −(x2 − x1 )(x3 − x1 )(x3 − x2 ).
x23 x3 1
The general formula is
x1n−1 x1n−2 · · ·
xn−1 xn−2 · · ·
2
2
det .
..
..
..
.
.
n−1
n−2
xn
xn
···
x21
x22
..
.
x2n
x1 1
Y
x2 1
=
±
(xi − xj ),
.. ..
. .
i>j
xn 1
where ± = (−1)n(n−1)/2 . It can be proved by induction using the same strategy as we used for
the 3× 3 case. The product on the right is the product of all differences xi − xj . This product
is non-zero, since we are assuming that all the points xi are distinct. Thus the Vandermonde
matrix is invertible, and a solution to the Lagrange interpolation problem always exists.
Now let’s use MATLAB/Octave to see how this interpolation works in practice.
We begin by putting some points xi into a vector X and the corresponding points yi into a
vector Y.
>X=[0 0.2 0.4 0.6 0.8 1.0]
>Y=[1 1.1 1.3 0.8 0.4 1.0]
We can use the plot command in MATLAB/Octave to view these points. The command
plot(X,Y) will pop open a window and plot the points (xi , yi ) joined by straight lines. In
this case we are not interested in joining the points (at least not with straight lines) so we
add a third argument: ’o’ plots the points as little circles. (For more information you can
type help plot on the MATLAB/Octave command line.) Thus we type
>plot(X,Y,’o’)
>axis([-0.1, 1.1, 0, 1.5])
>hold on
The axis command adjusts the axis. Normally when you issue a new plot command, the
existing plot is erased. The hold on prevents this, so that subsequent plots are all drawn on
the same graph. The original behaviour is restored with hold off.
When you do this you should see a graph appear that looks something like this.
22
I.2 Interpolation
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Now let’s compute the interpolation polynomial. Luckily there are build in functions in
MATLAB/Octave that make this very easy. To start with, the function vander(X) returns
the Vandermonde matrix corresponding to the points in X. So we define
>V=vander(X)
V =
0.00000
0.00032
0.01024
0.07776
0.32768
1.00000
0.00000
0.00160
0.02560
0.12960
0.40960
1.00000
0.00000
0.00800
0.06400
0.21600
0.51200
1.00000
0.00000
0.04000
0.16000
0.36000
0.64000
1.00000
0.00000
0.20000
0.40000
0.60000
0.80000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
We saw above that the coefficients of the interpolation polynomial are given by the solution
a to the equation V a = y. We find those coefficients using
>a=V\Y’
Let’s have a look at the interpolating polynomial. The MATLAB/Octave function polyval(a,X)
takes a vector X of x values, say x1 , x2 , . . . xk and returns a vector containing the values
p(x1 ), p(x2 ), . . . p(xk ), where p is the polynomial whose coefficients are in the vector a, that
is,
p(x) = a1 xn−1 + a2 xn−2 + · · · + an−1 x + an
So plot(X,polyval(a,X)) would be the command we want, except that with the present
definition of X this would only plot the polynomial at the interpolation points. What we
want is to plot the polynomial for all points, or at least for a large number. The command
linspace(0,1,100) produces a vector of 100 linearly spaced points between 0 and 1, so the
following commands do the job.
>XL=linspace(0,1,100);
>YL=polyval(a,XL);
>plot(XL,YL);
>hold off
23
I Linear Equations
The result looks pretty good
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
The MATLAB/Octave commands for this example are in lagrange.m.
Unfortunately, things get worse when we increase the number of interpolation points. One
clue that there might be trouble ahead is that even for only six points the condition number
of V is quite high (try it!). Let’s see what happens with 18 points. We will take the x
values to be equally spaced between 0 and 1. For the y values we will start off by taking
yi = sin(2πxi ). We repeat the steps above.
>X=linspace(0,1,18);
>Y=sin(2*pi*X);
>plot(X,Y,’o’)
>axis([-0.1 1.1 -1.5 1.5])
>hold on
>V=vander(X);
>a=V\Y’;
>XL=linspace(0,1,500);
>YL=polyval(a,XL);
>plot(XL,YL);
The resulting picture looks okay.
1.5
1
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
But look what happens if we change one of the y values just a little. We add 0.02 to the
fifth y value, redo the Lagrange interpolation and plot the new values in red.
24
I.2 Interpolation
>Y(5) = Y(5)+0.02;
>plot(X(5),Y(5),’or’)
>a=V\Y’;
>YL=polyval(a,XL);
>plot(XL,YL,’r’);
>hold off
The resulting graph makes a wild excursion and even though it goes through the given points,
it would not be a satisfactory interpolating function in a practical situation.
1.5
1
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
A calculation reveals that the condition number is
>cond(V)
ans =
1.8822e+14
If we try to go to 20 points equally spaced between 0 and 1, the Vandermonde matrix is so
ill conditioned that MATLAB/Octave considers it to be singular.
25
I Linear Equations
I.2.3 Cubic splines
In the last section we saw that Lagrange interpolation becomes impossible to use in practice
if the number of points becomes large. Of course, the constraint we imposed, namely that the
interpolating function be a polynomial of low degree, does not have any practical basis. It is
simply mathematically convenient. Let’s start again and consider how ship and airplane designers actually drew complicated curves before the days of computers. Here is a picture of a
draughtsman’s spline (taken from http://pages.cs.wisc.edu/~deboor/draftspline.html
where you can also find a nice photo of such a spline in use)
It consists of a bendable but stiff strip held in position by a series of weights called ducks.
We will try to make a mathematical model of such a device.
We begin again with points (x1 , y1 ), (x2 , y2 ), . . . (xn , yn ) in the plane. Again we are looking
for a function f (x) that goes through all these points. This time, we want to find the function
that has the same shape as a real draughtsman’s spline. We will imagine that the given points
are the locations of the ducks.
Our first task is to identify a large class of functions that represent possible shapes for the
spline. We will write down three conditions for a function f (x) to be acceptable. Since the
spline has no breaks in it the function f (x) should be continuous. Moreover f (x) should pass
through the given points.
Condition 1: f (x) is continuous and f (xi ) = yi for i = 1, . . . , n.
The next condition reflects the assumption that the strip is stiff but bendable. If the strip
were not stiff, say it were actually a rubber band that just is stretched between the ducks,
then our resulting function would be a straight line between each duck location (xi , yi ). At
each duck location there would be a sharp bend in the function. In other words, even though
the function itself would be continuous, the first derivative would be discontinuous at the
duck locations. We will interpret the words “bendable but stiff” to mean that the first
derivatives of f (x) exist. This leads to our second condition.
26
I.2 Interpolation
Condition 2: The first derivative f ′ (x) exists and is continuous everywhere, including each
interior duck location xi .
In between the duck locations we will assume that f (x) is perfectly smooth and that higher
derivatives behave nicely when we approach the duck locations from the right or the left.
This leads to
Condition 3: For x in between the duck points xi the higher order derivatives f ′′ (x), f ′′′ (x), . . .
all exist and have left and right limits as x approaches each xi .
In this condition we are allowing for the possibility that f ′′ (x) and higher order derivatives
have a jump at the duck locations. This happens if the left and right limits are different.
The set of functions satisfying conditions 1, 2 and 3 are all the possible shapes of the spline.
How do we decide which one of these shapes is the actual shape of the spline? To do this we
need to invoke a bit of the physics of bendable strips. The bending energy E[f ] of a strip
whose shape is described by the function f is given by the integral
Z xn
2
f ′′ (x) dx
E[f ] =
x1
The actual spline will relax into the shape that makes E[f ] as small as possible. Thus, among
all the functions satisfying conditons 1, 2 and 3, we want to choose the one that minimizes
E[f ].
This minimization problem is similiar to ones considered in calculus courses, except that
instead of real numbers, the variables in this problem are functions f satisfying conditons 1,
2 and 3. In calculus, the minimum is calculated by “setting the derivative to zero.” A similar
procedure is described in the next section. Here is the result of that calculation: Let F (x)
be the function describing the shape that makes E[f ] as small as possible. In other words,
• F (x) satisfies condtions 1, 2 and 3.
• If f (x) also satisfies conditions 1, 2 and 3, then E[F ] ≤ E[f ].
Then, in addition to conditions 1, 2 and 3, F (x) satisfies
Condition a: In each interval (xi , xi+1 ), the function F (x) is a cubic polynomial. In other
words, for each interval there are coefficients Ai , Bi , Ci and Di such that F (x) =
Ai x3 + Bi x2 + Ci x + Di for all x between xi and xi+1 . The coefficients can be different
for different intervals.
Condition b: The section derivative F ′′ (x) is continuous.
Condition c: When x is an endpoint (either x1 or xn ) then F ′′ (x) = 0
As we will see, there is exactly one function satisfying conditions 1, 2, 3, a, b and c.
27
I Linear Equations
I.2.4 The minimization procedure
In this section we explain the minimization procedure leading to a mathematical description
of the shape of a spline. In other words, we show that if among all functions f (x) satisfying
conditions 1, 2 and 3, the function F (x) is the one with E[f ] the smallest, then F (x) also
satisfies conditions a, b and c.
The idea is to assume that we have found F (x) and then try to deduce what properties it
must satisfy. There is actually a is a hidden assumption here — we are assuming that the
minimizer F (x) exists. This is not true for every minimization problem (think of minimizing
the function (x2 +1)−1 for −∞ < x < ∞). However the spline problem does have a minimizer,
and we will leave out the step of proving it exists.
Given the minimizer F (x) we want to wiggle it a little and consider functions of the form
F (x) + ǫh(x), where h(x) is another function and ǫ be a number. We want to do this in such
a way that for every ǫ, the function F (x) + ǫh(x) still satisfies conditions 1, 2 and 3. Then
we will be able to compare E[F ] with E[F + ǫh]. A little thought shows that functions of
form F (x) + ǫh(x) will satsify conditions 1, 2 and 3 for every value of ǫ if h satisfies
Condition 1’: h(xi ) = 0 for i = 1, . . . , n.
together with conditions 2 and 3 above.
Now, the minimization property of F says that each fixed function h satisfying 1’, 2 and 3
the function of ǫ given by E[F + ǫh] has a local minimum at ǫ = 0. From Calculus we know
that this implies that
dE[F + ǫh] = 0.
(I.1)
dǫ
ǫ=0
Now we will actually compute this derivative with respect to ǫ and see what information
we can get from the fact that it is zero for every choice of h(x) satisfying conditions 1’, 2
and 3. To simplify the presentation we will assume that there are only three points (x1 , y1 ),
(x2 , y2 ) and (x3 , y3 ). The goal of this computation is to establish that equation (??) can be
rewritten as (??).
To begin, we compute
Z x3
d(F ′′ (x) + ǫh′′ (x))2 dE[F + ǫh] =
dx
0=
dǫ
dǫ
x1
ǫ=0
ǫ=0
Z x3
2 (F ′′ (x) + ǫh′′ (x))h′′ (x)ǫ=0 dx
=
x1
Z x3
F ′′ (x)h′′ (x)dx
=2
Z x3
Zx1x2
′′
′′
F ′′ (x)h′′ (x)dx
F (x)h (x)dx + 2
=2
x1
28
x2
I.2 Interpolation
We divide by 2 and integrate by parts in each integral. This gives
Z
Z x2
x=x3
x=x2
′′′
′
′′
′
′′
′
F (x)h (x)dx + F (x)h (x) x=x2 −
0 = F (x)h (x) x=x1 −
x1
x3
F ′′′ (x)h′ (x)dx
x2
In each boundary term we have to take into account the possibility that F ′′ (x) is not continuous across the points xi . Thus we have to use the appropriate limit from the left or the
right. So, for the first boundary term
x=x
F ′′ (x)h′ (x)x=x21 = F ′′ (x2 −)h′ (x2 ) − F ′′ (x1 +)h′ (x1 )
Notice that since h′ (x) is continuous across each xi we need not distinguish the limits from
the left and the right. Expanding and combining the boundary terms we get
0 = −F ′′ (x1 +)h′ (x1 ) + F ′′ (x2 −) − F ′′ (x2 +) h′ (x2 ) + F ′′ (x3 −)h′ (x3 )
Z x3
Z x2
′′′
′
F ′′′ (x)h′ (x)dx
F (x)h (x)dx −
−
x1
x2
Now we integrate by parts again. This time the boundary terms all vanish because h(xi ) =
0 for every i. Thus we end up with the equation
0 = −F ′′ (x1 +)h′ (x1 ) + F ′′ (x2 −) − F ′′ (x2 +) h′ (x2 ) + F ′′ (x3 −)h′ (x3 )
Z x3
Z x2
′′′′
F ′′′′ (x)h(x)dx
(I.2)
F (x)h(x)dx −
+
x1
x2
as desired.
Recall that this equation has to be true for every choice of h satisfying conditions 1’, 2
and 3. For different choices of h(x) we can extract different pieces of information about the
minimizer F (x).
To start, we can choose h that
R x is zero everywhere except in the open interval (x1 , x2 ). For
all such h we then obtain 0 = x12 F ′′′′ (x)h(x)dx. This can only happen if
F ′′′′ (x) = 0 for
x1 < x < x2
Thus we conclude that the fourth derivative F ′′′′ (x) is zero in the interval (x1 , x2 ).
Once we know that F ′′′′ (x) = 0 in the interval (x1 , x2 ), then by integrating both sides we
can conclude that F ′′′ (x) is constant. Integrating again, we find F ′′ (x) is a linear polynomial.
By integrating four times, we see that F (x) is a cubic polynomial in that interval. When
doing the integrals, we must not extend the domain of integration over the boundary point
x2 since F ′′′′ (x) may not exist (let alone by zero) there.
Similarly F ′′′′ (x) must also vanish in the interval (x2 , x3 ), so F (x) is a (possibly different)
cubic polynomial in the interval (x2 , x3 ).
29
I Linear Equations
(An aside: to understand better why the polynomials might be different in the intervals
(x1 , x2 ) and (x3 , x4 ) consider the function g(x) (unrelated to the spline problem) given by
(
0 for x1 < x < x2
g(x) =
1 for x2 < x < x3
Then g′ (x) = 0 in each interval, and an integration tells us that g is constant in each interval.
However, g′ (x2 ) does not exist, and the constants are different.)
We have established that F (x) satisfies condition a.
Now that we know that F ′′′′ (x) vanishes in each interval, we can return to (??) and write
it as
0 = −F ′′ (x1 +)h′ (x1 ) + F ′′ (x2 −) − F ′′ (x2 +) h′ (x2 ) + F ′′ (x3 −)h′ (x3 )
Now choose h(x) with h′ (x1 ) = 1 and h′ (x2 ) = h′ (x3 ) = 0. Then the equation reads
F ′′ (x1 +) = 0
Similarly, choosing h(x) with h′ (x3 ) = 1 and h′ (x1 ) = h′ (x2 ) = 0 we obtain
F ′′ (x3 −) = 0
This establishes condition c.
Finally choosing h(x) with h′ (x2 ) = 1 and h′ (x1 ) = h′ (x3 ) = 0 we obtain
F ′′ (x2 −) − F ′′ (x2 +) = 0
In other words, F ′′ must be continuous across the interior duck position. Thus shows that
condition b holds, and the derivation is complete.
This calculation is easily generalized to the case where there are n duck positions x1 , . . . , xn .
A reference for this material is Essentials of numerical analysis, with pocket calculator
demonstrations, by Henrici.
I.2.5 The linear equations for cubic splines
Let us now turn this description into a system of linear equations. In each interval (xi , xi+1 ),
for i = 1, . . . n − 1, f (x) is given by a cubic polynomial pi (x) which we can write in the form
pi (x) = ai (x − xi )3 + bi (x − xi )2 + ci (x − xi ) + di
for coefficients ai , bi , ci and di to be determined. For each i = 1, . . . n − 1 we require that
pi (xi ) = yi and pi (xi+1 ) = yi+1 . Since pi (xi ) = di , the first of these equations is satisfied if
di = yi . So let’s simply make that substitution. This leaves the n − 1 equations
pi (xi+1 ) = ai (xi+1 − xi )3 + bi (xi+1 − xi )2 + ci (xi+1 − xi ) + yi = yi+1 .
30
I.2 Interpolation
Secondly, we require continuity of the first derivative across interior xi ’s. This translates to
p′i (xi+1 ) = p′i+1 (xi+1 ) or
3ai (xi+1 − xi )2 + 2bi (xi+1 − xi ) + ci = ci+1
for i = 1, . . . , n − 2, giving an additional n − 2 equations. Next, we require continuity of the
second derivative across interior xi ’s. This translates to p′′i (xi+1 ) = p′′i+1 (xi+1 ) or
6ai (xi+1 − xi ) + 2bi = 2bi+1
for i = 1, . . . , n − 2, once more giving an additional n − 2 equations. Finally, we require that
p′′1 (x1 ) = p′′n−1 (xn ) = 0. This yields two more equations
2b1 = 0
6an−1 (xn − xn−1 ) + 2bn−1 = 0
for a total of 3(n − 1) equations for the same number of variables.
We now specialize to the case where the distances between the points xi are equal. Let
L = xi+1 − xi be the common distance. Then the equations read
ai L3 + bi L2
2
3ai L + 2bi L
+ci L
+ci
6ai L + 2bi
−2bi+1
= yi+1 − yi
− ci+1
=0
=0
for i = 1 . . . n − 2 together with
an−1 L3 + bn−1 L2
+cn−1 L
= yn − yn−1
+ 2b1
=0
6an−1 L + 2bn−1
=0
We make one more simplification. After multiplying some of the equations with suitable
powers of L we can write these as equations for αi = ai L3 , βi = bi L2 and γi = ci L. They
have a very simple block structure. For example, when n = 4 the matrix form of the equations
is
1 1 1 0 0
0 0 0
0
α1
y2 − y1
3 2 1 0 0 −1 0 0
0
β1 0
6 2 0 0 −2 0 0 0
0 γ1 0
0 0 0 1 1
1 0 0
0
α2 y3 − y2
0 0 0 3 2
1 0 0 −1
β2 = 0
0 0 0 6 2
0 0 −2 0 γ2 0
0 0 0 0 0
0 1 1
1
α3 y4 − y3
0 2 0 0 0
0 0 0
0 β3 0
0 0 0 0
0
0
6
2
0
γ3
0
Notice that the matrix in this equation does not depend on the points (xi , yi ). It has a 3 × 3
31
I Linear Equations
block structure. If we define the 3 × 3 blocks
1 1 1
N = 3 2 1
6 2 0
0 0
0
M = 0 0 −1
0 −2 0
0 0 0
0 = 0 0 0
0 0 0
0 0 0
T = 0 2 0
0 0 0
1 1 1
V = 0 0 0
6 2 0
then the matrix in our equation has the form
N M
S = 0 N
T 0
0
M
V
Once we have solved the equation for the coefficients αi , βi and γi , the function F (x) in the
interval (xi , xi+1 ) is given by
F (x) = pi (x) = αi
x − xi
L
3
+ βi
x − xi
L
2
+ γi
x − xi
L
+ yi
Now let us use MATLAB/Octave to plot a cubic spline. To start, we will do an example
with four interpolation points. The matrix S in the equation is defined by
>N=[1 1 1;3 2 1;6 2 0];
>M=[0 0 0;0 0 -1; 0 -2 0];
>Z=zeros(3,3);
>T=[0 0 0;0 2 0; 0 0 0];
>V=[1 1 1;0 0 0;6 2 0];
>S=[N M Z; Z N M; T Z V]
S =
1
3
32
1
2
1
1
0
0
0
0
0
-1
0
0
0
0
0
0
I.2 Interpolation
6
0
0
0
0
0
0
2
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
3
6
0
0
0
-2
1
2
2
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
0
6
0
0
0
-2
1
0
2
0
0
-1
0
1
0
0
Here we used the function zeros(n,m) which defines an n × m matrix filled with zeros.
To proceed we have to know what points we are trying to interpolate. We pick four (x, y)
values and put them in vectors. Remember that we are assuming that the x values are
equally spaced.
>X=[1, 1.5, 2, 2.5];
>Y=[0.5, 0.8, 0.2, 0.4];
We plot these points on a graph.
>plot(X,Y,’o’)
>hold on
Now let’s define the right side of the equation
>b=[Y(2)-Y(1),0,0,Y(3)-Y(2),0,0,Y(4)-Y(3),0,0];
and solve the equation for the coefficients.
>a=S\b’;
Now let’s plot the interpolating function in the first interval. We will use 50 closely spaced
points to get a smooth looking curve.
>XL = linspace(X(1),X(2),50);
Put the first set of coefficients (α1 , β1 , γ1 , y1 ) into a vector
>p = [a(1) a(2) a(3) Y(1)];
33
I Linear Equations
Now we put the values p1 (x) into the vector YL. First we define the values (x − x1 )/L and put
them in the vector XLL. To get the values x − x1 we want to subtract the vector with X(1)
in every position from X. The vector with X(1) in every position can be obtained by taking
a vector with 1 in every position (in MATLAB/Octave this is obtained using the function
ones(n,m)) and multiplying by the number X(1). Then we divide by the (constant) spacing
between the xi values.
>L = X(2)-X(1);
>XLL = (XL - X(1)*ones(1,50))/L;
Now we evaluate the polynomial p1 (x) and plot the resulting points.
>YL = polyval(p,XLL);
>plot(XL,YL);
To complete the plot, we repeat this steps for the intervals (x2 , x3 ) and (x3 , x4 ).
>XL = linspace(X(2),X(3),50);
>p = [a(4) a(5) a(6) Y(2)];
>XLL = (XL - X(2)*ones(1,50))/L;
>YL = polyval(p,XLL);
>plot(XL,YL);
>XL = linspace(X(3),X(4),50);
>p = [a(7) a(8) a(9) Y(3)];
>XLL = (XL - X(3)*ones(1,50))/L;
>YL = polyval(p,XLL);
>plot(XL,YL);
The result looks like this:
34
I.2 Interpolation
0.8
0.7
0.6
0.5
0.4
0.3
0.2
1
1.2
1.4
1.6
1.8
2
2.2
2.4
I have automated the procedure above and put the result in two files splinemat.m and
plotspline.m. splinemat(n) returns the 3(n−1)×3(n−1) matrix used to compute a spline
through n points while plotspline(X,Y) plots the cubic spline going through the points in
X and Y. If you put these files in you MATLAB/Octave directory you can use them like this:
>splinemat(3)
ans =
1
3
6
0
0
0
1
2
2
0
2
0
1
1
0
0
0
0
0
0
0
1
0
6
0
0
-2
1
0
2
0
-1
0
1
0
0
and
>X=[1, 1.5, 2, 2.5];
>Y=[0.5, 0.8, 0.2, 0.4];
>plotspline(X,Y)
35
I Linear Equations
to produce the plot above.
Let’s use these functions to compare the cubic spline interpolation with the Lagrange
interpolation by using the same points as we did before. Remember that we started with the
points
>X=linspace(0,1,18);
>Y=sin(2*pi*X);
Let’s plot the spline interpolation of these points
>plotspline(X,Y);
Here is the result with the Lagrange interpolation added (in red). The red (Lagrange) curve
covers the blue one and its impossible to tell the curves apart.
1.5
1
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
Now we move one of the points slightly, as before.
>Y(5) = Y(5)+0.02;
Again, plotting the spline in blue and the Lagrange interpolation in red, here are the results.
36
I.2 Interpolation
1.5
1
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
This time the spline does a much better job! Let’s check the condition number of the
matrix for the splines. Recall that there are 18 points.
>cond(splinemat(18))
ans =
32.707
Recall the Vandermonde matrix had a condition number of 1.8822e+14. This shows that
the system of equations for the splines is very much better conditioned, by 13 orders of
magnitude!!
Code for splinemat.m and plotspline.m
function S=splinemat(n)
L=[1 1 1;3 2 1;6 2 0];
M=[0 0 0;0 0 -1; 0 -2 0];
Z=zeros(3,3);
T=[0 0 0;0 2 0; 0 0 0];
V=[1 1 1;0 0 0;6 2 0];
S=zeros(3*(n-1),3*(n-1));
for k=[1:n-2]
37
I Linear Equations
for l=[1:k-1]
S(3*k-2:3*k,3*l-2:3*l) = Z;
end
S(3*k-2:3*k,3*k-2:3*k) = L;
S(3*k-2:3*k,3*k+1:3*k+3) = M;
for l=[k+2:n-1]
S(3*k-2:3*k,3*l-2:3*l) = Z;
end
end
S(3*(n-1)-2:3*(n-1),1:3)=T;
for l=[2:n-2]
S(3*(n-1)-2:3*(n-1),3*l-2:3*l) = Z;
end
S(3*(n-1)-2:3*(n-1),3*(n-1)-2:3*(n-1))=V;
end
function plotspline(X,Y)
n=length(X);
L=X(2)-X(1);
S=splinemat(n);
b=zeros(1,3*(n-1));
for k=[1:n-1]
b(3*k-2)=Y(k+1)-Y(k);
b(3*k-1)=0;
b(3*k)=0;
end
a=S\b’;
npoints=50;
XL=[];
YL=[];
for k=[1:n-1]
XL = [XL linspace(X(k),X(k+1),npoints)];
p = [a(3*k-2),a(3*k-1),a(3*k),Y(k)];
XLL = (linspace(X(k),X(k+1),npoints) - X(k)*ones(1,npoints))/L;
YL = [YL polyval(p,XLL)];
end
plot(X,Y,’o’)
38
I.2 Interpolation
hold on
plot(XL,YL)
hold off
I.2.6 Summary of MATLAB/Octave commands used in this section
How to access elements of a vector
a(i) returns the i-th element of the vector a
How to create a vector with linearly spaced elements
linspace(x1,x2,n) generates n points between the values x1 and x2.
How to create a matrix by concatenating other matrices
C= [A B] takes two matrices A and B and creates a new matrix C by concatenating A and
B horizontally
Other specialized matrix functions
zeros(n,m) creates a n-by-m matrix filled with zeros
ones(n,m) creates a n-by-m matrix filled with ones
vander(X) creates the Vandermonde matrix corresponding to the points in the vector X.
Note that the columns of the Vandermonde matrix are powers of the vector X.
Other useful functions and commands
polyval(a,X) takes a vector X of x values and returns a vector containing the values of
a polynomial p evaluated at the x values. The coefficients of the polynomial p (in
descending powers) are the values in the vector a.
sin(X) takes a vector X of values x and returns a vector containing the values of the function
sin x
plot(X,Y) plots vector Y versus vector X. Points are joined by a solid line. To change line
types (solid, dashed, dotted, etc.) or plot symbols (point, circle, star, etc.), include an
additional argument. For example, plot(X,Y,’o’) plots the points as little circle.
39
I Linear Equations
I.3 Finite difference approximations
Prerequisites and Learning Goals
From your work in previous courses, you should be able to
• explain what it is meant by a boundary value problem.
After completing this section, you should be able to
• Take a second order linear boundary value problem and write down the corresponding
finite difference equation.
• Use the finite difference equation and MATLAB/Octave to compute an approximate
solution.
• Use the MATLAB/Octave command diag.
• Describe the action of . (period) before a MATLAB/Octave operator.
I.3.1 Introduction and example
One of the most important applications of linear algebra is the approximate solution of
differential equations. In a differential equation we are trying to solve for an unknown
function. The basic idea is to turn a differential equation into a system of N × N linear
equations. As N becomes large, the vector solving the system of linear equations becomes a
better and better approximation to the function solving the differential equation.
In this section we will learn how to use linear algebra to find approximate solutions to a
boundary value problem of the form
f ′′ (x) + q(x)f (x) = r(x) for
0≤x≤1
subject to boundary conditions
f (0) = A,
f (1) = B.
This is a differential equation where the unknown quantity to be found is a function f (x).
The functions q(x) and r(x) are given (known) functions.
As differential equations go, this is a very simple one. For one thing it is an ordinary
differential equation (ODE), because it only involves one independent variable x. But the
finite difference methods we will introduce can also be applied to partial differential equations
(PDE).
It can be useful to have a picture in your head when thinking about an equation. Here is
a situation where an equation like the one we are studying arises. Suppose we want to find
the shape of a stretched hanging cable. The cable is suspended above the points x = 0 and
x = 1 at heights of A and B respectively and hangs above the interval 0 ≤ x ≤ 1. Our goal
is to find the height f (x) of the cable above the ground at every point x between 0 and 1.
40
I.3 Finite difference approximations
u(x)
A
B
0
1
x
The loading of the cable is described by a function 2r(x) that takes into account both the
weight of the cable and any additional load. Assume that this is a known function. The
height function f (x) is the function that minimizes the sum of the stretching energy and the
gravitational potential energy given by
Z 1
[(f ′ (x))2 + 2r(x)f (x)]dx
E[f ] =
0
subject to the condition that f (0) = A and f (1) = B. An argument similar (but easier) to
the one we did for splines shows that the minimizer satisfies the differential equation
f ′′ (x) = r(x).
So we end up with the special case of our original equation where q(x) = 0. Actually,
this special case can be solved by simply integrating twice and adjusting the constants of
integration to ensure f (0) = A and f (1) = B. For example, when r(x) = r is constant and
A = B = 1, the solution is f (x) = 1 − rx/2 + rx2 /2. We can use this exact solution to
compare against the approximate solution that we will compute.
I.3.2 Discretization
In the finite difference approach to solving differential equations approximately, we want to
approximate a function by a vector containing a finite number of sample values. Pick equally
spaced points xk = k/N , k = 0, . . . , N between 0 and 1. We will represent a function f (x)
by its values fk = f (xk ) at these points. Let
f0
f1
F = . .
..
fN
41
I Linear Equations
f(x)
f0 f1 f2 f3 f4 f5 f6 f7 f8
x0 x1 x2 x3 x4 x5 x6 x7 x8
x
At this point we throw away all the other information about the function, keeping only the
values at the sampled points.
f(x)
f0 f1 f2 f3 f4 f5 f6 f7 f8
x0 x1 x2 x3 x4 x5 x6 x7 x8
x
If this is all we have to work with, what should we use as an approximation to f ′ (x)? It
seems reasonable to use the slopes of the line segments joining our sampled points.
f(x)
f0 f1 f2 f3 f4 f5 f6 f7 f8
x0 x1 x2 x3 x4 x5 x6 x7 x8
x
Notice, though, that there is one slope for every interval (xi , xi+1 ) so the vector containing
the slopes has one fewer entry than the vector F. The formula for the slope in the interval
42
I.3 Finite difference approximations
(xi , xi+1 ) is (fi+1 − fi )/∆x where the distance ∆x = xi+1 − xi (in this case ∆x = 1/N ).
Thus the vector containing the slopes is
f0
f1 − f0
−1 1
0 0 ··· 0
f2 − f1
0 −1 1 0 · · · 0 f1
f2
0 −1 1 · · · 0
F′ = (∆x)−1 f3 − f2 = (∆x)−1 0
f3 = (∆x)−1 DN F
..
..
..
..
.. . .
..
.
. .
.
.
.
.
...
fN − fN −1
0
0
0 0 ··· 1
fN
where DN is the N × (N + 1) finite difference matrix in the formula above. The vector F′ is
our approximation to the first derivative function f ′ (x).
To approximate the second derivative f ′′ (x), we repeat this process to define the vector F′′ .
There will be one entry in this vector for each adjacent pair of slopes, that is, each adjacent
pair of entries of F′ . These are naturally labelled by the interior points x1 , x2 , . . . , xn−1 .
Thus we obtain
f0
1 −2 1
0 ··· 0 0 0
0 1 −2 1 · · · 0 0 0 f1
f2
1 −2 · · · 0 0 0
F′′ = (∆x)−2 DN −1 DN F = (∆x)−2 0 0
f3 .
..
..
..
..
.
.
.
..
..
..
..
.
. ..
.
.
.
.
0 0
0
0 · · · 1 −2 1
fN
Let rk = r(xk ) be the sampled points for the load function r(x) and define the vector
approximation for r at the interior points
r1
r = ... .
rN −1
The reason we only define this vector for interior points is that that is where F′′ is defined.
Now we can write down the finite difference approximation to f ′′ (x) = r(x) as
(∆x)−2 DN −1 DN F = r
or
DN −1 DN F = (∆x)2 r
This is a system of N − 1 equations in N + 1 unknowns. To get a unique solution, we need
two more equations. That is where the boundary conditions come in! We have two boundary
conditions, which in this case can simply be written as f0 = A and fN = B. Combining these
43
I Linear Equations
with the N − 1 equations for the interior
1 0
0
0 ···
1 −2 1
0 ···
0 1 −2 1 · · ·
0 0
1 −2 · · ·
..
.
..
..
..
..
.
.
.
.
0 0
0
0 ···
0 0
0
0 ···
points, we may rewrite the system of equations as
0 0 0
A
0 0 0
(∆x)2 r1
0 0 0
(∆x)2 r2
0 0 0 F =
.
..
..
..
..
.
.
.
.
2
(∆x) rN −1
1 −2 1
B
0 0 1
Note that it is possible to incorporate other types of boundary conditions by simply changing
the first and last equations.
Let’s define L to be the (N + 1) × (N + 1) matrix of coefficients for this equation, so that
the equation has the form
LF = b.
The first thing to do is to verify that L is invertible, so that we know that there is
a unique solution to the equation. It is not too difficult to compute the determinant
if you recall that the elementary row operations that add a multiple of one row to another do not change the value of the determinant. Using only this type of elementary
row operation, we can reduce L to an upper triangular matrix whose diagonal entries are
1, −2, −3/2, −4/3, −5/4, . . . , −N/(N − 1), 1. The determinant is the product of these entries,
and this equals ±N . Since this value is not zero, the matrix L is invertible.
It is worthwhile pointing out that a change in boundary conditions (for example, prescribing
the values of the derivative f ′ (0) and f ′ (1) rather than f (0) and f (1)) results in a different
matrix L that may fail to be invertible.
We should also ask about the condition number of L to determine how large the relative
error of the solution can be. We will compute this using MATLAB/Octave below.
Now let’s use MATLAB/Octave to solve this equation. We will start with the test case
where r(x) = 1 and A = B = 1. In this case we know that the exact solution is f (x) =
1 − x/2 + x2 /2.
We will work with N = 50. Notice that, except for the first and last rows, L has a constant
value of −2 on the diagonal, and a constant value of 1 on the off-diagonals immediately above
and below.
Before proceeding, we introduce the MATLAB/Octave command diag. For any vector D,
diag(D) is a diagonal matrix with the entries of D on the diagonal. So for example
>D=[1 2 3 4 5];
>diag(D)
44
I.3 Finite difference approximations
ans =
1
0
0
0
0
0
2
0
0
0
0
0
3
0
0
0
0
0
4
0
0
0
0
0
5
An optional second argument offsets the diagonal. So, for example
>D=[1 2 3 4];
>diag(D,1)
ans =
0
0
0
0
0
1
0
0
0
0
0
2
0
0
0
0
0
3
0
0
0
0
0
4
0
0
0
0
0
4
0
0
0
0
0
>diag(D,-1)
ans =
0
1
0
0
0
0
0
2
0
0
0
0
0
3
0
Now returning to our matrix L we can define it as
>N=50;
>L=diag(-2*ones(1,N+1)) + diag(ones(1,N),1) + diag(ones(1,N),-1);
>L(1,1) = 1;
>L(1,2) = 0;
>L(N+1,N+1) = 1;
>L(N+1,N) = 0;
The condition number of L for N = 50 is
45
I Linear Equations
>cond(L)
ans = 1012.7
We will denote the right side of the equation by b. To start, we will define b to be (∆x)2 r(xi )
and then adjust the first and last entries to account for the boundary values. Recall that
r(x) is the constant function 1, so its sampled values are all 1 too.
>dx = 1/N;
>b=ones(N+1,1)*dx^2;
>A=1; B=1;
>b(1) = A;
>b(N+1) = B;
Now we solve the equation for F.
>F=L\b;
The x values are N + 1 equally spaced points between 0 and 1,
>X=linspace(0,1,N+1);
Now we plot the result.
>plot(X,F)
1
0.98
0.96
0.94
0.92
0.9
0.88
0
46
0.2
0.4
0.6
0.8
1
I.3 Finite difference approximations
Let’s superimpose the exact solution in red.
>hold on
>plot(X,ones(1,N+1)-X/2+X.^2/2,’r’)
(The . before an operator tells MATLAB/Octave to apply that operator element by element,
so X.^2 returns an array with each element the corresponding element of X squared.)
1
0.98
0.96
0.94
0.92
0.9
0.88
0
0.2
0.4
0.6
0.8
1
The two curves are indistinguishable.
What happens if we increase the load at a single point? Recall that we have set the loading
function r(x) to be 1 everywhere. Let’s increase it at just one point. Adding, say, 5 to one of
the values of r is the same as adding 5(∆x)2 to the right side b. So the following commands
do the job. We are changing b11 which corresponds to changing r(x) at x = 0.2.
>b(11) = b(11) + 5*dx^2;
>F=L\b;
>hold on
>plot(X,F);
Before looking at the plot, let’s do this one more time, this time making the cable really
heavy at the same point.
47
I Linear Equations
>b(11) = b(11) + 50*dx^2;
>F=L\b;
>hold on
>plot(X,F);
Here is the resulting plot.
1
0.95
0.9
0.85
0.8
0.75
0
0.2
0.4
0.6
0.8
1
So far we have only considered the case of our equation f ′′ (x) + q(x)f (x) = r(x) where
q(x) = 0. What happens when we add the term containing q? We must sample the function
q(x) at the interior points and add the corresponding vector. Since we multiplied the equations for the interior points by (∆x)2 we must do the same to these terms. Thus we must
add the term
0 0 0 0 ··· 0
0
0
0
0 q1 0 0 · · · 0
0
0
q1 f1
0 0 q2 0 · · · 0
0
0
q2 f2
2 0
0
0
q
·
·
·
0
0
0
3
(∆x)2
=
(∆x)
F.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
. .
.
.
.
.
qN −1 fN −1
0 0 0 0 · · · 0 qN −1 0
0
0 0 0 0 ··· 0
0
0
In other words, we replace the matrix L in our equation with L + (∆x)2 Q where Q is the
(N + 1) × (N + 1) diagonal matrix with the interior sampled points of q(x) on the diagonal.
48
I.3 Finite difference approximations
I’ll leave it to a homework problem to incorporate this change in a MATLAB/Octave
calculation. One word of caution: the matrix L by itself is always invertible (with reasonable
condition number). However L + (∆x)2 Q may fail to be invertible. This reflects the fact
that the original differential equation may fail to have a solution for some choices of q(x) and
r(x).
I.3.3 Another example: the heat equation
In the previous example involving the loaded cable there was only one independent variable,
x, and as a result we ended up with an ordinary differential equation which determined
the shape. In this example we will have two independent variables, time t, and one spatial
dimension x. The quantities of interest can now vary in both space and time. Thus we
will end up with a partial differential equation which will describe how the physical system
behaves.
Imagine a long thin rod (a one-dimensional rod) where the only important spatial direction
is the x direction. Given some initial temperature profile along the rod and boundary conditions at the ends of the rod, we would like to determine how the temperature, T = T (x, t),
along the rod varies over time.
Consider a small section of the rod between x and x + ∆x. The rate of change of internal
energy, Q(x, t), in this section is proportional to the heat flux, q(x, t), into and out of the
section. That is
∂Q
(x, t) = −q(x + ∆x, t) + q(x, t).
∂t
Now the internal energy is related to the temperature by Q(x, t) = ρCp ∆xT (x, t), where ρ
and Cp are the density and specific heat of the rod (assumed here to be constant). Also,
from Fourier’s law, the heat flux through a point in the rod is proportional to the (negative)
temperature gradient at the point, i.e., q(x, t) = −K0 ∂T (x, t)/∂x, where K0 is a constant
(the thermal conductivity); this basically says that heat “flows” from hotter to colder regions.
Substituting these two relations into the above energy equation we get
∂T
∂T
∂T
(x, t) = K0
(x + ∆x, t) −
(x, t)
ρCp ∆x
∂t
∂x
∂x
⇒
∂T
K0
(x, t) =
∂t
ρCp
∂T
∂x (x
+ ∆x, t) −
∆x
∂T
∂x (x, t)
.
Taking the limit as ∆x goes to zero we obtain
∂2T
∂T
(x, t) = k 2 (x, t),
∂t
∂x
where k = K0 /ρCp is a constant. This partial differential equation is known as the heat
equation and describes how the temperature along a one-dimensional rod evolves.
49
I Linear Equations
We can also include other effects. If there was a temperature source or sink, S(x, t), then
this will contribute to the local change in temperature:
∂2T
∂T
(x, t) = k 2 (x, t) + S(x, t).
∂t
∂x
And if we also allow the rod to cool down along its length (because, say, the surrounding air
is a different temperature than the rod), then the differential equation becomes
∂2T
∂T
(x, t) = k 2 (x, t) − HT (x, t) + S(x, t),
∂t
∂x
where H is a constant (here we have assumed that the surrounding air temperature is zero).
In certain cases we can think about what the steady state of the rod will be. That is after
sufficiently long time (so that things have had plenty of time for the heat to “move around”
and for things to heat up/cool down), the temperature will cease to change in time. Once
this steady state is reached, things become independent of time, and the differential equation
becomes
∂2T
0 = k 2 (x) − HT (x) + S(x),
∂x
which is of the same form as the ordinary differential equation that we considered at the
start of this section.
50
Chapter II
Subspaces, Bases and Dimension
51
II Subspaces, Bases and Dimension
II.1 Subspaces, basis and dimension
Prerequisites and Learning Goals
From your work in previous courses, you should be able to
• Write down a vector as a linear combination of a set of vectors.
• Define linear independence for a collection of vectors.
• Define a basis for a vector subspace.
After completing this section, you should be able to
• Know the definitions of vector addition and scalar multiplication for vector spaces of
functions
• Decide whether a given collection of vectors forms a subspace.
• Recast the dependence or independence of a collection of vectors in Rn or Cn as a
statement about existence of solutions to a system of linear equations.
• Decide if a collection of vectors are dependent or independent.
• Define the span of a collection of vectors; show that given a set of vectors v1 , . . . , vk
the span span(v1 , . . . , vk ) is a subspace.
• Describe the significance of the two parts (independence and span) of the definition of
a basis.
• Check if a collection of vectors is a basis.
• Show that any basis for a subspace has the same number of elements.
• Show that any set of k linearly independent vectors v1 , . . . , vk in a k dimensional
subspace S is a basis of S.
• Define the dimension of a subspace.
52
II.1 Subspaces, basis and dimension
II.1.1 Vector spaces and subspaces
x1
In your previous linear algebra course, and for most of this course, vectors are n-tuples ...
xn
n
n
of numbers, either real or complex. The sets of all n-tuples, denoted R or C , are examples
of vector spaces.
In more advanced applications vector spaces of functions often occur. For example, an
electrical signal can be thought of as a real valued function x(t) of time t. If two signals x(t)
and y(t) are superimposed, the resulting signal is the sum that has the value x(t) + y(t) at
time t. This motivates the definition of vector addition for functions: the vector sum of the
functions x and y is the new function x + y defined by (x + y)(t) = x(t) + y(t). Similarly, if
s is a scalar, the scalar multiple sx is defined by (sx)(t) = sx(t). If you think of t as being a
continuous index, these definitions mirror the componentwise definitions of vector addition
and scalar multiplication for vectors in Rn or Cn .
It is possible to give an abstract definition of a vector space as any collection of objects
(the vectors) that can be added and multiplied by scalars, provided the addition and scalar
multiplication rules obey a set of rules. We won’t follow this abstract approach in this course.
A collection of vectors V contained in a given vector space is called a subspace if vector
addition and scalar multiplication of vectors in V stay in V . In other words, for any vectors
v1 , v2 ∈ V and any scalars c1 and c2 , the vector c1 v1 + c2 v2 lies in V too.
In three dimensional space R3 , examples of subspaces are lines and planes through the
origin. If we add or scalar multiply two vectors lying on the same line (or plane) the resulting
vector remains on the same line (or plane). Additional examples of subspaces are the trivial
subspace, containing the single vector 0, as well as the whole space itself.
Here is another example of a subspace. The set of n × n matrices can be thought of as
an n2 dimensional vector space. Within this vector space, the set of symmetric matrices
(satisfying AT = A) is a subspace. To see this, suppose A1 and A2 are symmetric. Then,
using the linearity property of the transpose, we see that
(c1 A1 + c2 A2 )T = c1 AT1 + c2 AT2 = c1 A1 + c2 A2
which shows that c1 A1 + c2 A2 is symmetric too.
We have encountered subspaces of functions in the section on interpolation. In Lagrange
interpolation we considered the set of all polynomials of degree at most m. This is a subspace
of the space of functions, since adding two polynomials of degree at most m results in another
polynomial, again of degree at most m, and scalar multiplication of a polynomial of degree
at most m yields another polynomial of degree at most m.
Another example of a subspace of functions is the set of all functions y(t) that satisfy the
differential equation y ′′ (t) + y(t) = 0. To check that this is a subspace, we must verify that
if y1 (t) and y2 (t) both solve the differential equation, then so does c1 y1 (t) + c2 y2 (t) for any
choice of scalars c1 and c2 .
53
II Subspaces, Bases and Dimension
II.1.2 Linear dependence and independence
To begin we review the definition of linear dependence and independence. A linear combination of vectors v1 , . . . , vk is a vector of the form
k
X
i=1
ci vi = c1 v1 + c2 v2 + · · · + ck vk
for some choice of numbers c1 , c2 , . . . , ck .
The vectors v1 , . . . , vk are called linearly dependent P
if there exist numbers c1 , c2 , . . . , ck
that are not all zero, such that the linear combination ki=1 ci vi = 0
On the other hand, the vectors are called linearly independent if the only linear combination
of the vectors equaling zero has every ci = 0. In other words
k
X
i=1
ci vi = 0 implies
c1 = c2 = · · · = ck = 0
7
1
1
For example, the vectors 1, 0 and 1 are linearly dependent because
7
1
1
1
1
7
0
1 1 + 6 0 − 1 1 = 0
1
1
7
0
If v1 , . . . , vk are linearly dependent, then at least one of the vi ’s can be written as a linear
combination of the others. To see this suppose that
c1 v1 + c2 v2 + · · · + ck vk = 0
with not all of the ci ’s zero. Then we can solve for any of the vi ’s whose coefficient ci is not
zero. For instance, if c1 is not zero we can write
v1 = −(c2 /c1 )v2 − (c3 /c1 )v3 − · · · − (ck /c1 )vk
This means any linear combination we can make with the vectors v1 , . . . , vk can be achieved
without using v1 , since we can simply replace the occurrence of v1 with the expression on
the right.
Sometimes it helps to have a geometrical picture. In three dimensional space R3 , three
vectors are linearly dependent if they lie in the same plane.
The columns of a matrix in echelon form are linearly independent if and only if every
column is a pivot column. We illustrate this with two examples.
54
II.1 Subspaces, basis and dimension
1 ∗
The matrix 0 2
0 0
pivot column. Here ∗
∗
∗ is an example of a matrix in echelon form where each column is a
3
denotes an arbitrary entry.
To see that the columns are linearly independent suppose that
1
∗
∗
0
c1 0 + c2 2 + c3 ∗ = 0
0
0
3
0
Then, equating the bottom entries we find 3c3 = 0 so c3 = 0. But once we know c3 = 0 then
the equation reads
0
∗
1
c1 0 + c2 2 = 0
0
0
0
which implies that c2 = 0 too, and similarly c1 = 0
Similarly, for a matrix in echelon form (even if, as in the example below, it is not completely
reduced), the pivot columns are linearly independent. For example the first, second and fifth
columns in the matrix
1 1 1 1 0
0 1 2 5 5
0 0 0 0 1
are independent. However, the non-pivot columns can be written as linear combination of
the pivot columns. For example
1
1
1
2 = − 0 + 2 1
0
0
0
so if there are non-pivot columns, then the set of all columns is linearly dependent. This is
particularly easy to see for a matrix in reduced row echelon form, like
1 0 1 1 0
0 1 2 5 0 .
0 0 0 0 1
In this case the pivot columns are standard basis vectors (see below), which are obviously
independent. It is easy to express the other columns as linear combinations of these.
Recall that for a matrix U in echelon form, the presence or absence of non-pivot columns
determines whether the homogeneous equation U x = 0 has any non-zero solutions. By the
discussion above, we can say that the columns of a matrix U in echelon form are linearly
dependent exactly when the homogeneous equation U x = 0 has a non-zero solution.
In fact, this is true for any matrix. Suppose that the vectors v1 , . . . , vk are the columns of
a matrix A so that
A = v1 v2 · · · vk .
55
II Subspaces, Bases and Dimension
If we put the coefficients c1 , c2 , . . . , ck into a vector
c1
c2
c=.
..
ck
then
Ac = c1 v1 + c2 v2 + · · · + ck vk
is the linear combination of the columns v1 , . . . , vk with coefficients ci .
Now it follows directly from the definition of linear dependence that the columns of A are
linearly dependent if there is a non-zero solution c to the homogeneous equation
Ac = 0
On the other hand, if the only solution to the homogeneous equation is c = 0 then the
columns v1 , . . . , vk are linearly independent.
To compute whether a given collection of vectors is dependent or independent we can place
them in the columns of a matrix A and reduce to echelon form. If the echelon form has only
pivot columns, then the vectors are independent. On the other hand, if the echelon form has
some non-pivot columns, then the equation Ac = 0 has some non-zero solutions and so the
vectors are dependent.
Let’s try this with the vectors in the example above in MATLAB/Octave.
>V1=[1
>V2=[1
>V3=[7
>A=[V1
1 1]’;
0 1]’;
1 7]’;
V2 V3]
A =
1
1
1
1
0
1
7
1
7
>rref(A)
ans =
1
0
0
0
1
0
1
6
0
Since the third column is a non-pivot column, the vectors are linearly dependent.
56
II.1 Subspaces, basis and dimension
II.1.3 Span
Given a collection of vectors v1 , . . . , vk we may form a subspace of all possible linear combinations. This subspace is called span(v1 , . . . , vk ) or the space spanned by the vi ’s. It is a subspace because if we start with any two elements of span(v1 , . . . , vk ), say c1 v1 +c2 v2 +· · ·+ck vk
and d1 v1 + d2 v2 + · · · + dk vk then a linear combination of these linear combinations is again
a linear combination since
s1 (c1 v1 + c2 v2 + · · · + ck vk ) + s2 (d1 v1 + d2 v2 + · · · + dk vk ) =
(s1 c1 + s2 d1 )v1 + (s1 c2 + s2 d2 )v2 + · · · + (s1 ck + s2 dk )vk
1
0
0
For example the span of the three vectors 0 , 1 and 0 is the whole three dimensional
0
0
1
space,
because
every
vector
is
a
linear
combination
of
these.
The span of the four vectors
1
0
0
1
0, 1, 0 and 1 is the same.
1
1
0
0
II.1.4 Basis
A collection of vectors v1 , . . . , vk contained in a subspace V is called a basis for that subspace
if
1. span(v1 , . . . , vk ) = V , and
2. v1 , . . . , vk are linearly independent.
Condition (1) says that any vector in V can be written as a linear combination of v1 , . . . , vk .
Condition (2) says that there is exactly one way of doing this. Here is the argument. Suppose
there are two ways of writing the same vector v ∈ V as a linear combination:
v = c1 v1 + c2 v2 + · · · + ck vk
v = d1 v1 + d2 v2 + · · · + dk vk
Then by subtracting these equations, we obtain
0 = (c1 − d1 )v1 + (c2 − d2 )v2 + · · · + (ck − dk )vk
Linear independence now says that every coefficient in this sum must be zero. This implies
c1 = d1 , c2 = d2 . . . ck = dk .
57
II Subspaces, Bases and Dimension
Example: Rn has the standard basis e1 , e2 , . . . , en where
1
0
0
1
e1 = e2 = · · ·
..
..
.
.
1
1
Another basis for
is
,
. To see this, notice that saying that any vector y can
1
−1
1
1
+ c2
is the same as saying that the equation
be written in a unique way as c1
1
−1
R2
c1
1 1
=x
1 −1 c2
always has a unique solution. This is true.
A basis for the vector space P2 of polynomials of degree at most two is given by {1, x, x2 }.
These polynomials clearly span P2 since every polynomial p ∈ P2 can be written as a linear
combination p(x) = c0 · 1 + c1 x + c2 x2 . To show independence, suppose that c0 · 1 + c1 x + c2 x2
is the zero polynomial. This means that c0 · 1 + c1 x + c2 x2 = 0 for every value of x. Taking
the first and second derivatives of this equation yields that c1 + 2c2 x = 0 and 2c2 = 0 for
every value of x. Substituting x = 0 into each of these equations we find c0 = c1 = c2 = 0.
2
Notice that if
we
represent the polynomial p(x) = c0 · 1 + c1 x + c2 x ∈ P2 by the vector
c0
of coefficients c1 ∈ R3 , then the vector space operations in P2 are mirrored perfectly in
c2
R3 . In other words, adding or scalar multiplying polynomials in P2 is the same as adding or
scalar multiplying the corresponding vectors of coefficients in R3 .
This sort of correspondence can be set up whenever we have a basis v1 , v2 , . . . , vk for a
vector space V . In this case every vector v has a unique representation
c1 v1 +c2 v2 +· · ·+ck vk
c1
c2
and we can represent the vector v ∈ V by the vector . ∈ Rk (or Ck ). In some sense this
..
ck
says that we can always think of finite dimensional vector spaces as being copies of Rn or
Cn . The only catch is that the the correspondence that gets set up between vectors in V and
vectors in Rn or Cn depends on the choice of basis.
It is intuitively clear that, say, a plane in three dimensions will always have a basis of two
vectors. Here is an argument that shows that any two bases for a subspace S of Rk or Ck
will always have the same number of elements. Let v1 , . . . , vn and w1 , . . . , wm be two bases
58
II.1 Subspaces, basis and dimension
for a subspace S. Let’s try to show that n must be the same as m. Since the vi ’s span V we
can write each wi as a linear combination of vi ’s. We write
wj =
n
X
ai,j vi
i=1
for each j = 1, . . . , m. Let’s put all the coefficients into an n × m matrix A = [ai,j ]. If we
form the matrix k ×m matrix W = [w1 |w2 | · · · |wm ] and the k ×n matrix V = [v1 |v2 | · · · |vm ]
then the equation above can be rewritten
W =VA
1
1
1
4
To understand this construction consider the two bases
0 , −1
and
2 , −2
−1
0
−6
1
3
for
a subspace in R (in fact this subspace is the plane through the origin with normal vector
1
1). Then we may write
1
4
1
1
2 = 6 0 − 2 −1
−6
−1
0
1
1
1
−2 = − 0 + 2 −1
1
−1
0
and the equation W = V A for this example reads
4
1
1
1 2 −2 = 0 −1 6 −1
−2 2
−6 1
−1 0
Returning now to the general case, suppose that m > n. Then A has more columns than
rows. So its echelon form must have some non-pivot columns which implies that there must
be some non-zero solution to Ac = 0. Let c 6= 0 be such a solution. Then
W c = V Ac = 0
But this is impossible because the columns of W are linearly dependent. So it can’t be true
that m > n. Reversing the roles of V and W we find that n > m is impossible too. So it
must be that m = n.
We have shown that any basis for a subspace S has the same number of elements. Thus
it makes sense to define the dimension of a subspace S to be the number of elements in any
59
II Subspaces, Bases and Dimension
basis for S.
Here is one last fact about bases: any set of k linearly independent vectors {v1 , . . . , vk } in
a k dimensional subspace S automatically spans S and is therefore a basis. To see this (in
the case that S is a subspace of Rn or Cn ) we let {w1 , . . . , wk } be a basis for S, which also
will have k elements. Form V = [v1 | · · · |vk ] and W = [w1 | · · · |wk ]. Then the construction
above gives V = W A for a k ×k matrix A. The matrix A must be invertible. Otherwise there
would be a non-zero solution c to Ac = 0. This would imply V c = W Ac = 0 contradicting
the independence of the rows of V . Thus we can write W = V A−1 which shows that every
wk is a linear combination of vi ’s.
This shows that the vi ’s must span S because every vector in S is a linear combination of
the basis vectors wk ’s which in turn are linear combinations of the vi ’s.
As an example of this, consider again the space P2 of polynomials of degree at most 2.
We claim that the polynomials {1, (x − a), (x − a)2 } (for any constant a) form a basis. We
already know that the dimension of this space is 3, so we only need to show that these three
polynomials are independent. The argument for that is almost the same as before.
II.2 The four fundamental subspaces for a matrix
From your work in previous courses, you should be able to
• Recognize and use the property of transposes for which (AB)T = B T AT for any matrices A and B.
• Define the inner (dot) product of two vectors, and its properties (symmetry, linearity),
and explain its geometrical meaning.
• Use the inner product to decide if two vectors are orthogonal, and to compute the angle
between two vectors.
• State the Cauchy-Schwarz inequality and know for which vectors the inequality is an
equality.
After completing this section, you should be able to
• Define the four fundamental subspaces N (A), R(A), N (AT ), and R(AT ), associated to
a matrix A and its transpose AT .
• Express the Gaussian elimination process performed to reduce a matrix A to its row
reduced echelon form matrix U as a matrix factorization, A = EU , using elementary
matrices, and be able to perform the steps using MATLAB/Octave.
• Compute bases for each of the four fundamental subspaces N (A), R(A), N (AT ) and
R(AT ) of a matrix A.
60
II.2 The four fundamental subspaces for a matrix
• Be able to compute the rank of a matrix.
• State the formulas for the dimension of each of the four subspaces and be able to explain
why they are true.
• Explain what it means for two subspaces to be orthogonal (V ⊥ W ) and for one
subspace to be the orthogonal complement of another (V = W ⊥ ).
• State which of the fundamental subspaces are orthogonal to each other and explain
why, verify the orthogonality relations in examples, and use the orthogonality relation
for R(A) to test whether the equation Ax = b has a solution.
• Use MATLAB/Octave to compute the inner product of two vectors and the angle
between them.
• Be familiar with the MATLAB/Octave function eye().
II.2.1 Nullspace N(A) and Range R(A)
There are two important subspaces associated to any matrix. Let A be an n × m matrix. If
x is m dimensional, then Ax makes sense and is a vector in n dimensional space.
The first subspace associated to A is the nullspace (or kernel) of A denoted N (A) (or
Ker(A)). It is defined as all vectors x solving the homogeneous equation for A, that is
N (A) = {x : Ax = 0}
This is a subspace because if Ax1 = 0 and Ax2 = 0 then
A(c1 x1 + c2 x2 ) = c1 Ax1 + c2 Ax2 = 0 + 0 = 0.
The nullspace is a subspace of m dimensional space Rm .
The second subspace is the range (or column space) of A denoted R(A) (or C(A)). It is
defined as all vectors of the form Ax for some x. From our discussion above, we see that
R(A) is the the span (or set of all possible linear combinations) of its columns. This explains
the name “column space”. The range is a subspace of n dimensional space Rn .
The four fundamental subspaces for a matrix are the nullspace N (A) and range R(A) for
A together with the nullspace N (AT ) and range R(AT ) for the transpose AT .
61
II Subspaces, Bases and Dimension
II.2.2 Finding basis and dimension of N(A)
Example: Let
1 3 3 10
A = 2 6 −1 −1 .
1 3 1
4
To calculate a basis for the nullspace N (A) and determine its dimension we need to find the
solutions to Ax = 0. To do this we first reduce A to reduced row echelon form U and solve
U x = 0 instead, since this has the same solutions as the original equation.
>A=[1 3 3 10;2 6 -1 -1;1 3 1 4];
>rref(A)
ans =
1
0
0
3
0
0
0
1
0
1
3
0
x1
x2
This means that x =
x3 is in N (A) if
x4
x
1 3 0 1 1
0
0 0 1 3 x2 = 0
x3
0 0 0 0
0
x4
We now divide the variables into basic variables, corresponding to pivot columns, and free
variables, corresponding to non-pivot columns. In this example the basic variables are x1
and x3 while the free variables are x2 and x4 . The free variables are the parameters in the
solution. We can solve for the basic variables in terms of the free ones, giving x3 = −3x4
and x1 = −3x2 − x4 . This leads to
x1
−3x2 − x4
−3
−1
x2
x
1
2
=
0
x3 −3x4 = x2 0 + x4 −3
x4
x4
0
1
−3
−1
1
0
The vectors
0 and −3 span the nullspace since every element of N (A) is a linear
0
1
combination of them. They are also linearly independent because if the linear combination
62
II.2 The four fundamental subspaces for a matrix
on the right of the equation above is zero, then by looking at the second entry of the vector (corresponding to the first free variable) we find x2 = 0 and looking at the last entry
(corresponding to the second free variable) we find x4 = 0. So both coefficients must be zero.
To find a basis for N (A) in general we first compute U = rref(A) and determine which
variables are basic and which are free. For each free variable we form a vector as follows.
First put a 1 in the position corresponding to that free variable and a zero in every other
free variable position. Then fill in the rest of the vector in such a way that U x = 0. (This
is easy to do!) The set all such vectors - one for each free variable - is a basis for N (A).
II.2.3 The matrix version of Gaussian elimination
How are a matrix A and its reduced row echelon form U = rref(A) related? If A and U are
n × m matrices, then there exists an invertible n × n matrix such that
A = EU
E −1 A = U
This immediately explains why the N (A) = N (U ), because if Ax = 0 then U x = E −1 Ax = 0
and conversely if Ax = 0 then U x = EAx = 0.
What is this matrix E? It can be thought of as a matrix record of the Gaussian elimination
steps taken to reduce A to U . It turns out performing an elementary row operation is the
same as multiplying on the left by an invertible square matrix. This invertible square matrix,
called an elementary matrix, is obtained by doing the row operation in question to the identity
matrix.
Suppose we start with the matrix
>A=[1 3 3 10;2 6 -1 -1;1 3 1 4]
A =
1
2
1
3
6
3
3
-1
1
10
-1
4
The first elementary row operation that we want to do is to subtract twice the first row
from the second row. Let’s do this to the 3 × 3 identity matrix I (obtained with eye(3) in
MATLAB/Octave) and call the result E1
>E1 = eye(3)
E1 =
1
0
0
0
1
0
0
0
1
63
II Subspaces, Bases and Dimension
>E1(2,:) = E1(2,:)-2*E1(1,:)
E1 =
1
-2
0
0
1
0
0
0
1
Now if we multiply E1 and A we obtain
>E1*A
ans =
1
0
1
3
0
3
3
-7
1
10
-21
4
which is the result of doing that elementary row operation to A. Let’s do one more step.
The second row operation we want to do is to subtract the first row from the third. Thus
we define
>E2 = eye(3)
E2 =
1
0
0
0
1
0
0
0
1
>E2(3,:) = E2(3,:)-E2(1,:)
E2 =
1
0
-1
0
1
0
0
0
1
and we find
64
II.2 The four fundamental subspaces for a matrix
>E2*E1*A
ans =
1
0
0
3
0
0
3
-7
-2
10
-21
-6
which is one step further along in the Gaussian elimination process. Continuing in this way
until we eventually arrive at U so that
Ek Ek−1 · · · E2 E1 A = U
−1
Thus A = EU with E = E1−1 E2−1 · · · Ek−1
Ek−1 . For the example above it turns out that
1 3
−6
E = 2 −1 −18
1 1
−9
which we can check:
>A=[1 3 3 10;2 6 -1 -1;1 3 1 4]
A =
1
2
1
3
6
3
3
-1
1
10
-1
4
>U=rref(A)
U =
1
0
0
3
0
0
0
1
0
1
3
0
>E=[1 3 -6; 2 -1 -18; 1 1 -9];
>E*U
ans =
1
2
1
3
6
3
3
-1
1
10
-1
4
65
II Subspaces, Bases and Dimension
If we do a partial elimination then at each step we can write A = E ′ U ′ where U ′ is the
resulting matrix at the point we stopped, and E ′ is obtained from the Gaussian elimination
step up to that point. A common place to stop is when U ′ is in echelon form, but the entries
above the pivots have not been set to zero. If we can achieve this without doing any row
swaps along the way then E ′ turns out to be lower triangular matrix. Since U ′ is upper
triangular, this is called the LU decomposition of A.
II.2.4 A basis for R(A)
The ranges or column spaces R(A) and R(U ) are not the same in general, but they are
related. In fact, the vectors in R(A) are exactly all the vectors in R(U ) multiplied by E,
where E is the invertible matrix in the equation A = EU . We can write this relationship as
R(A) = ER(U )
To see this notice that if x ∈ R(U ), that is, x = U y for some y then Ex = EU y = Ay is in
R(A). Conversely if x ∈ R(A), that is, x = Ay for some y then x = EE −1 Ay = EU y so x
is E times a vector in R(U ).
Now if we can find a basis u1 , u2 , . . . , uk for R(U ), the vectors Eu1 , Eu2 , . . . , Euk form a
basis for R(A). (Homework exercise)
But a basis for the column space R(U ) is easy to find. They are exactly the pivot columns
of U . If we multiply these by E we get a basis for R(A). But if
#
" " #
A = a1 a2 · · · am , U = u1 u2 · · · um
then the equation A = EU can be written
# "
#
" a1 a2 · · · am = Eu1 Eu2 · · · Eum
From this we see that the columns of A that correspond to pivot columns of U form a basis
for R(A). This implies that the dimension of R(A) is the number of pivot columns in U .
II.2.5 The rank of a matrix
We define the rank of the matrix A, denoted r(A) to be the number of pivot columns of U .
Then we have shown that for an n × m matrix A
dim(R(A)) = r(A)
dim(N (A)) = m − r(A)
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II.2 The four fundamental subspaces for a matrix
II.2.6 Bases for R(AT ) and N(AT )
Of course we could find R(AT ) and N (AT ) by computing the reduced row echelon form for
AT and following the steps above. But then we would miss an important relation between
the dimensions of these spaces.
Let’s start with the column space R(AT ). The columns of AT are the rows of A (written
as column vectors instead of row vectors). So R(AT ) is the row space of A.
It turns out that R(AT ) and R(U T ) are the same. This follows from A = EU . To see this
take the transpose of this equation. Then AT = U T E T . Now suppose that x ∈ R(AT ). This
means that x = AT y for some y. But then x = U T E T y = U T y′ where y′ = E T y so x ∈
R(U T ). Similarly, if x = U T y for some y then x = U T E T (E T )−1 y = AT (E T )−1 y = AT y′
for y′ = (E T )−1 y. So every vector in R(U T ) is also in R(AT ). Here we used that E and
hence E T is invertible.
Now we know that R(AT ) = R(U T ) is spanned by the columns of U T . But since U T is
in reduced row echelon form, its non-zero columns are independent. Therefore, the non-zero
columns of U T form a basis for R(AT ). There is one of these for every pivot. This leads to
dim(R(AT )) = r(A) = dim(R(A))
The final subspace to consider is N (AT ). From our work above we know that
dim(N (AT )) = n − dim(R(AT )) = n − r(A).
Finding a basis is trickier. It might be easiest to find the reduced row echelon form of AT .
But if we insist on using A = EU or AT = U T E T we could proceed by multiplying on the
right be the inverse of E T . This gives
AT (E T )−1 = U T
Now notice that the last n − r(A) columns of U T are zero, since U is in reduced row echelon
form. So the last n − r(A) columns of (E T )−1 are in the the nullspace of AT . They also have
to be independent, since (E T )−1 is invertible.
Thus the last n − r(A) of (E T )−1 form a basis for N (AT ).
From a practical point of view, this is not so useful since we have to compute the inverse
of a matrix. It might be just as easy to reduce AT . (Actually, things are slightly better if
we use the LU decomposition. The same argument shows that the last n − r(A) columns of
(LT )−1 also form a basis for N (AT ). But LT is an upper triangular matrix, so its inverse is
faster to compute.)
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II Subspaces, Bases and Dimension
II.2.7 Orthogonal vectors and subspaces
In preparation for our discussion of the orthogonality relations for the fundamental subspaces
of matrix we review some facts about orthogonal vectors and subspaces.
Recall that the dot product, or inner product of two vectors
x1
y1
x2
y2
x= . y= .
..
..
xn
yn
is denoted by x · y or hx, yi and defined by
xT y = x1 x2 · · ·
y1
n
y2 X
xn . =
xi yi
..
yn
i=1
Some important properties of the inner product are symmetry
x·y =y·x
and linearity
(c1 x1 + c2 x2 ) · y = c1 x1 · y + c2 x2 · y.
The (Euclidean) norm, or length, of a vector is given by
v
u n
uX
√
x2i
kxk = x · x = t
i=1
An important property of the norm is that kxk = 0 implies that x = 0.
The geometrical meaning of the inner product is given by
x · y = kxkkyk cos(θ)
where θ is the angle between the vectors. The angle θ can take values from 0 to π.
The Cauchy–Schwarz inequality states
|x · y| ≤ kxkkyk.
It follows from the previous formula because | cos(θ)| ≤ 1. The only time that equality occurs
in the Cauchy–Schwarz inequality, that is x · y = kxkkyk, is when cos(θ) = ±1 and θ is either
0 or π. This means that the vectors are pointed in the same or in the opposite directions.
68
II.2 The four fundamental subspaces for a matrix
The vectors x and y are orthogonal if x · y = 0. Geometrically this means either that one
of the vectors is zero or that they are at right angles. This follows from the formula above,
since cos(θ) = 0 implies θ = π/2.
Another way to see that x · y = 0 means that vectors are orthogonal is from Pythagoras’
formula. If x and y are at right angles then kxk2 + kyk2 = kx + yk2 .
x+y
y
x
But kx + yk2 = (x + y) · (x + y) = kxk2 + kyk2 + 2x · y so Pythagoras’ formula holds
exactly when x · y = 0.
To compute the inner product of (column) vectors X and Y in MATLAB/Octave we use
the formula x · y = xT y. Thus the inner product can be computed using X’*Y. (If X and Y
are row vectors, the formula is X*Y’.)
The norm of a vector X is computed by norm(X). In MATLAB/Octave inverse trig functions
are computed with asin(), acos() etc. So the angle between column vectors X and Y could
be computed as
> acos(X’*Y/(norm(X)*norm(Y)))
Two subspaces V and W are said to be orthogonal if every vector in V is orthogonal to
every vector in W . In this case we write V ⊥ W .
W
V
T
S
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II Subspaces, Bases and Dimension
In this figure V ⊥ W and also S ⊥ T .
A related concept is the orthogonal complement. The orthogonal complement of V , denoted
V is the subspace containing all vectors orthogonal to V . In the figure W = V ⊥ but T 6= S ⊥
since T contains only some of the vectors orthogonal to S.
⊥,
If we take the orthogonal complement of V ⊥ we get back the original space V : This is
certainly plausible from the pictures. It is also obvious that V ⊆ (V ⊥ )⊥ , since any vector
in V is perpendicular to vectors in V ⊥ . If there were a vector in (V ⊥ )⊥ not contained in
V we could subtract its projection onto V (defined in the next chapter) and end up with a
non-zero vector in (V ⊥ )⊥ that is also in V ⊥ . Such a vector would be orthogonal to itself,
which is impossible. This shows that
(V ⊥ )⊥ = V.
One consequence of this formula is that V = W ⊥ implies V ⊥ = W . Just take the orthogonal
complement of both sides and use (W ⊥ )⊥ = W .
II.2.8 Orthogonality relations for the fundamental subspaces of a matrix
Let A be an n × m matrix. Then N (A) and R(AT ) are subspaces of Rm while N (AT ) and
R(A) are subspaces of Rn .
These two pairs of subspaces are orthogonal:
N (A) = R(AT )⊥
N (AT ) = R(A)⊥
We will show that the first equality holds for any A. The second equality then follows by
applying the first one to AT .
These relations are based on the formula
(AT x) · y = x · (Ay)
This formula follows from the product formula (AB)T = B T AT for transposes, since
(AT x) · y = (AT x)T y = xT (AT )T y = xT Ay = x · (Ay)
First, we show that N (A) ⊆ R(AT )⊥ . To do this, start with any vector x ∈ N (A). This
means that Ax = 0. If we compute the inner product of x with any vector in R(AT ), that
is, any vector of the form AT y, we get (AT y) · x = y · Ax = y · 0 = 0. Thus x ∈ R(AT )⊥ .
This shows N (A) ⊆ R(AT )⊥ .
Now we show the opposite inclusion, R(AT )⊥ ⊆ N (A). This time we start with x ∈
R(AT )⊥ . This means that x is orthogonal to every vector in R(AT ), that is, to every
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II.2 The four fundamental subspaces for a matrix
vector of the form AT y. So (AT y) · x = y · (Ax) = 0 for every y. Pick y = Ax. Then
(Ax) · (Ax) = kAxk2 = 0. This implies Ax = 0 so x ∈ N (A). We can conclude that
R(AT )⊥ ⊆ N (A).
These two inclusions establish that N (A) = R(AT )⊥ .
Let’s verify these orthogonality relations in
1
A= 1
2
Then
an example. Let
2 1 1
3 0 1
5 1 2
1 0
3 1
rref(A) = 0 1 −1 0
0 0
0 0
Thus we get
1
0
rref(AT ) =
0
0
0
1
0
0
1
1
0
0
−3
−1
1
, 0
N (A) = span
1 0
0
1
2
1
R(A) = span
1 , 3
2
5
−1
T
N (A ) = span −1
1
1
0
0 1
T
R(A ) = span ,
3
−1
1
0
We can now verify directly that every vector in the basis for N (A) is orthogonal to every
vector in the basis for R(AT ), and similarly for N (AT ) and R(A).
Does the equation
2
Ax = 1
3
have a solution? We
can use the ideas above to answer this question easily. We are really
2
asking whether 1 is contained in R(A). But, according to the orthogonality relations, this
3
71
II Subspaces, Bases and Dimension
2
is the same as asking whether 1 is contained in N (AT )⊥ . This is easy to check. Simply
3
compute the dot product
2
−1
1 · −1 = −2 − 1 + 3 = 0.
3
1
Since the result is zero, we conclude that a solution exists.
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II.3 Graphs and Networks
II.3 Graphs and Networks
Prerequisites and Learning Goals
From your work in previous courses you should be able to
• State Ohm’s law for a resistor.
• State Kirchhoff’s laws for a resistor network.
After completing this section, you should be able to
• Be able to write down the incidence matrix of a directed graph, and to draw the graph
given the incidence matrix.
• Define the Laplace operator or Laplacian for a graph and be able to write it down.
• When the edges of a graph represent resistors or batteries in a circuit, you should be
able to
– interpret each of the four subspaces associated with the incidence matrix and their
dimensions in terms of voltage and current vectors, and verify their orthogonality
relations.
– write down Ohm’s law for all the edges of the graph in matrix form using the
Laplacian.
– express the connection between Kirchoff’s law and the nullspace of the Laplacian.
– use the voltage-to-current map to calculate the voltages and currents in the network when a battery is attached .
– use the voltage-to-current map to calculate the effective resistance between two
nodes in the network.
• Re-order rows and columns of a matrix and extract submatrices in MATLAB/Octave.
II.3.1 Directed graphs and their incidence matrix
A directed graph is a collection of vertices (or nodes) connected by edges with arrows. Here
is a graph with 4 vertices and 5 edges.
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II Subspaces, Bases and Dimension
1
1
2
2
5
4
3
4
3
Graphs come up in many applications. For example, the nodes could represent computers
and the arrows internet connections. Or the nodes could be factories and the arrows represent
movement of goods. We will mostly focus on a single interpretation where the edges represent
resistors or batteries hooked up in a circuit.
In this interpretation we will be assigning a number to each edge to indicate the amount
of current flowing through that edge. This number can be positive or negative. The arrows
indicate the direction associated to a positive current.
The incidence matrix of a graph is an n × m matrix, where n is the number of edges and m
is the number of vertices. We label the rows by the edges in the graph and the columns by
the vertices. Each row of the matrix corresponds to an edge in the graph. It has a −1 in the
place corresponding to the vertex where the arrow starts and a 1 in the place corresponding
to the vertex where the arrow ends.
Here is the incidence matrix for the illustrated graph.
1
2
3
4
1 −1
1
0
0
2
1
0
0 −1
3 0
0 −1
1
4 0 −1
0
1
5
1
0
0 −1
The columns of the matrix have the following interpretation. The column representing
a given vertex has a +1 for each arrow coming in to that vertex and a −1 for each arrow
leaving the vertex.
Given an incidence matrix, the corresponding graph can easily be drawn. What is the
graph for
−1
1
0
0 −1
1?
1
0 −1
(Answer: a triangular loop.)
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II.3 Graphs and Networks
II.3.2 Nullspace and range of incidence matrix and its transpose
We now wish to give an interpretation of the fundamental subspaces associated with the
incidence matrix of a graph. Let’s call the matrix D. In our example D actsonvectors
v1
v2
v ∈ R4 and produces a vector Dv in R5 . We can think of the vector v =
v3 as an
v4
v2 − v1
v3 − v2
assignment of a voltage to each of the nodes in the graph. Then the vector Dv =
v4 − v3
v4 − v2
v1 − v4
assigns to each edge the voltage difference across that edge. The matrix D is similar to the
derivative matrix when we studied finite difference approximations. It can be thought of as
the derivative matrix for a graph.
II.3.3 The null space N(D)
This is the set of voltages v for which the voltage differences in Dv are all zero. This means
that any two nodes connected by an edge will have the same voltage. In our example,
this
1
1
implies all the voltages are the same, so every vector in N (D) is of the form v = s
1 for
1
1
1
some s. In other words, the null space is one dimensional with basis
1.
1
For a graph that has several disconnected pieces, Dv = 0 will force v to be constant on
each connected component of the graph. Each connected component will contribute one basis
vector to N (D). This is the vector that is equal to 1 on that component and zero everywhere
else. Thus dim(N (D)) will be equal to the number of disconnected pieces in the graph.
II.3.4 The range R(D)
The range of D consists of all vectors b in R5 that are voltage differences, i.e., b = Dv for
some v. We know that the dimension of R(D) is 4 − dim(N (D)) = 4 − 1 = 3. So the set of
voltage difference vectors must be restricted in some way. In fact a voltage difference vector
will have the property that the sum of the differences around a closed loop is zero. In the
75
II Subspaces, Bases and Dimension
b1
b2
example the edges 1,
4,
5 form a loop, so if b =
b3 is a voltage difference vector then
b4
b5
v2 − v1
v3 − v2
b1 + b4 + b5 = 0 We can check this directly in the example. Since b = Dv =
v4 − v3
v4 − v2
v1 − v4
we check that (v2 − v1 ) + (v4 − v2 ) + (v1 − v4 ) = 0. In the example graph there are three
loops, namely 1,
4,
5 and 2,
3,
4 and 1,
2,
3,
5 . The corresponding equations that
the components of a vector b must satisfy to be in the range of D are
b1 + b4 + b5 = 0
b2 + b3 − b4 = 0
b1 + b2 + b3 + b5 = 0
Notice the minus sign in the second equation corresponding to a backwards arrow. However
these equations are not all independent, since the third is obtained by adding the first two.
There are two independent equations that the components of b must satisfy. Since R(D) is
3 dimensional, there can be no additional constraints.
y1
y2
Now we wish to find interpretations for the null space and the range of DT . Let y =
y3
y4
y5
5
be a vector in R which we
interpret as being an assignment of a current to each edge in
y5 − y1
y
1 − y2 − y4
the graph. Then D T y =
y2 − y3 . This vector assigns to each node the amount of
y3 + y4 − y5
current collecting at that node.
II.3.5 The null space N(D T )
This is the set of current vectors y ∈ R5 which do not result in any current building up
(or draining away) at any of the nodes. We know that the dimension of this space must be
5 − dim(R(DT )) = 5 − dim(R(D)) = 5 − 3 = 2. We can guess at a basis for this space by
noting
that
current running around a loop will not build up at any of the nodes. The loop
1
0
vector
1,
4,
5. We can verify that this
0 represents a current running around the loop 1
1
76
II.3 Graphs and Networks
vector lies in the null space of D T :
1
0
−1 0
0
0
1
1 −1 0 −1 0 0 0
0 =
0
0
1 −1 0
0
1
0
0
0
1
1 −1
1
0
1
1
1
The current vectors corresponding to the other two loops are
1 and 1. However
−1
0
0
1
these three vectors are not linearly independent. Any choice of two of these vectors are
independent, and form a basis.
II.3.6 The range R(D T )
x1
x2
T
This is the set of vectors in R4 of the form
x3 = D y. With our interpretation these are
x4
vectors which measure how the currents in y are building up or draining away from each
node. Since the current that is building up at one node must have come from some other
nodes, it must be that
x1 + x2 + x3 + x4 = 0
In our example, this can be checked directly. This one condition in R4 results in a three
dimensional subspace.
II.3.7 Summary and Orthogonality relations
The two subspaces R(D) and N (DT ) are subspaces of R5 . The subspace N (DT ) contains all
linear combination of loop vectors, while R(D) contains all vectors whose dot product with
loop vectors is zero. This verifies the orthogonality relation R(D) = N (DT )⊥ .
The two subspaces N (D) and R(DT ) are subspaces of R4 . The subspace N (D) contains
constant vectors, while R(DT ) contains all vectors orthogonal to constant vectors. This
verifies the other orthogonality relation N (D) ⊥ R(DT ).
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II Subspaces, Bases and Dimension
II.3.8 Resistors and the Laplacian
Now we suppose that each edge of our graph represents a resistor. This means that we
associate with the ith edge a resistance Ri . Sometimes it is convenient to use conductances
γi which are defined to be the reciprocals of the resistances, that is, γi = 1/Ri .
R1
1
2
R5
R2
R
4
3
4
R3
If we begin by an assignment of voltage to every node, and put these numbers in a vector
v ∈ R4 . Then Dv ∈ R5 represents the vector of voltage differences for each of the edges.
Given the resistance Ri for each edge, we can now invoke Ohm’s law to compute the current
flowing through each edge. For each edge, Ohm’s law states that
(∆V )i = ji Ri ,
where (∆V )i is the voltage drop across the edge, ji is the current flowing through that edge,
and Ri is the resistance. Solving for the current we obtain
ji = Ri−1 (∆V )i .
Notice that the voltage drop (∆V )i in this formula is exactly the ith component of the vector
Dv. So if we collect all the currents flowing along each edge in a vector j indexed by the
edges, then Ohm’s law for all the edges can be written as
j = R−1 Dv
where
R1 0
0
0
0
0 R2 0
0
0
R=
0 R3 0
0
0
0
0
0 R4 0
0
0
0
0 R5
is the diagonal matrix with the resistances on the diagonal.
78
II.3 Graphs and Networks
Finally, if we multiply j by the matrix DT the resulting vector
J = DT j = D T R−1 Dv
has one entry for each node, representing the total current flowing in or out of that node
along the edges that connect to it.
The matrix
L = D T R−1 D
appearing in this formula is called the Laplacian. It is similar to the second derivative matrix
that appeared when we studied finite difference approximations.
One important property of the Laplacian is symmetry, that is the fact that LT = L. To
see this recall that the transpose of a product of matrices is the product of the transposes in
reverse order ((ABC)T = C T B T AT ). This implies that
T
LT = (D T R−1 D)T = DT R−1 D = L
Here we used that D T
T
T
= D and that R−1 , being a diagonal matrix, satisfies R−1 = R−1 .
Let’s determine the entries of L. To start we consider the case where all the resistances
have the same value 1 so that R = R−1 = I. In this case L = DT D. Let’s start with the
example graph above. Then
−1
1
0
0
−1
0
0
0
1
2
−1
0
−1
1
0
1 −1
0 −1
−1
0 −1
0
3 −1 −1
=
L=
0
0
−1
1
0 −1
0
1 −1
0
0
2 −1
0 −1
0
1
0
0
1
1 −1
−1 −1 −1
3
1
0
0 −1
Notice that the ith diagonal entry is the total number of edges connected to the ith node.
The i, j entry is −1 if the ith node is connected to the jth node, and 0 otherwise.
This pattern describes the Laplacian L for any graph. To see this, write
D = [d1 |d2 |d3 | · · · |dm ]
Then the i, j entry of DT D is dTi dj . Recall that di has an entry of −1 for every edge leaving
the ith node, and a 1 for every edge coming in. So dTi di , the diagonal entries of DT D, are
the sum of (±1)2 , with one term for each edge connected to the ith node. This sum gives
the total number of edges connected to the ith node. To see this in the example graph, let’s
consider the first node. This node has two edges connected to it and
−1
0
d1 =
0
0
1
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II Subspaces, Bases and Dimension
Thus the 1, 1 entry of the Laplacian is
dT1 d1 = (−1)2 + 12 = 2
On the other hand, if i 6= j then the vectors di and dj have a non-zero entry in the same
position only if one of the edges leaving the ith node is coming in to the jth node or vice
versa. For a graph with at most one edge connecting any two nodes (we usually assume this)
this means that dTi dj will equal −1 if the ith and jth nodes are connected by an edge, and
zero otherwise. For example, in the graph above the first edge leaves the first node, so that
d1 has a −1 in the first position. This first edge comes in to the second node so d2 has a
+1 in the first position. Otherwise, there is no overlap in these vectors, since no other edges
touch both these nodes. Thus
1
−1
dT1 d2 = −1 0 0 0 1
0 = −1
−1
0
What happens if the resistances are not all equal to one? In this case we must replace
D with R−1 D in the calculation above. This multiplies the kth row of D with γk = 1/Rk .
Making this change in the calculations above leads to the following prescription for calculating
the entries of L. The diagonal entries are given by
X
Li,i =
γk
k
Where the sum goes over all edges touching the ith node. When i 6= j then
(
−γk if nodes i and j are connected with edge k
Li,j =
0 if nodes i and j are not connected
II.3.9 Kirchhoff’s law and the null space of L
Kirchhoff’s law states that currents cannot build up at any node. If v is the voltage vector
for a circuit, then we saw that Lv is the vector whose ith entry is the total current building
up at the ith node. Thus, for an isolated circuit that is not hooked up to any batteries,
Kirchhoff’s law can be written as
Lv = 0
By definition, the solutions are exactly the vectors in the nullspace N (L) of L. It turns out
that N (L) is the same as N (D), which contains all constant voltage vectors. This is what
we should expect. If there are no batteries connected to the circuit the voltage will be the
same everywhere and no current will flow.
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II.3 Graphs and Networks
To see N (L) = N (D) we start with a vector v ∈ N (D). Then Dv = 0 implies Lv =
D T R−1 Dv = DT R−1 0 = 0. This show that v ∈ N (L) too, that is, N (D) ⊆ N (L)
To show the opposite inclusion we first note that the matrix R−1 can be factored into a
−1
−1/2 R−1/2 where R−1/2 is the diagonal matrix with
product of invertible
√ matrices R = R
diagonal entries 1/ Ri . This is possible because each Ri is a positive number. Also, since
R−1/2 is a diagonal matrix it is equal to its transpose, that is, R−1/2 = (R−1/2 )T .
Now suppose that Lv = 0. This can be written D T (R−1/2 )T R−1/2 Dv = 0. Now we
multiply on the left with vT . This gives
vT DT (R−1/2 )T R−1/2 Dv = (R−1/2 Dv)T R−1/2 Dv = 0
But for any vector w, the number wT w is the dot product of w with itself which is equal to
the length of w squared. Thus the equation above can be written
kR−1/2 Dvk2 = 0
This implies that R−1/2 Dv = 0. Finally, since R−1/2 is invertible, this yields Dv = 0. We
have shown that any vector in N (L) also is contained in N (D). Thus N (L) ⊆ N (D) and
together with the previous inclusion this yields N (L) = N (D).
II.3.10 Connecting a battery
To see more interesting behaviour in a circuit, we pick two nodes and connect them to a
battery. For example, let’s take our example circuit above and connect the nodes 1 and 2.
1
2
R1
R5
R2
R
4
3
4
R3
The terminals of a battery are kept at a fixed voltage. Thus the voltages v1 and v2 are
now known, say,
v1 = b1
v2 = b2
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II Subspaces, Bases and Dimension
Of course, it is only voltage differences that have physical meaning, so we could set b1 = 0.
Then b2 would be the voltage of the battery.
At the first and second nodes there now will be current flowing in and out from the battery.
Let’s call these currents J1 and J2 . At all the other nodes the total current flowing in and
out is still zero, as before.
How are the equations for the circuit modified? For simplicity let’s set all the resistances
Ri = 1. The new equations are
2 −1
0 −1
b1
J1
−1
3 −1 −1 b2 J2
=
0 −1
2 −1 v3 0
−1 −1 −1
3
v4
0
Two of the voltages v1 and v2 have changed their role in these equations from being unknowns
to being knowns. On the other hand, the first two currents, which were originally known
quantities (namely zero) are now unknowns.
Since the current flowing into the network should equal the current flowing out, weexpect
J1
J2
that J1 = −J2 . This follows from the orthogonality relations for L. The vector
0 is
0
contained in R(L). But R(L) = N (LT )⊥ = N (L)⊥ (since L = LT ). But we know that N (L)
consists of all constant vectors. Hence
J1
1
J2 1
· = J1 + J2 = 0
0 1
1
0
To solve this system of equations we write it in block matrix form
A BT b
J
=
B C
v
0
where
and
2 −1
A=
−1
3
b
b= 1
b2
0 −1
B=
−1 −1
v
v= 3
v4
2 −1
C=
−1
3
J
J= 1
J2
0
0=
0
Our system of equations can then be written as two 2 × 2 systems.
Ab + B T v = J
Bb + Cv = 0
82
II.3 Graphs and Networks
We can solve the second equation for v. Since C is invertible
v = −C −1 Bb
Using this value of v in the first equation yields
J = (A − B T C −1 B)b
The matrix A − B T C −1 B is the voltage-to-current map. In our example
1 −1
T −1
A − B C B = (8/5)
−1
1
In fact, for any circuit the voltage to current map is given by
1 −1
T −1
A−B C B =γ
−1
1
This can be
deduced from two facts: (i) A−B T C −1 B is symmetric and (ii) R(A−B T C −1 B) =
1
span
. You are asked to carry this out in a homework problem.
−1
Notice that this form of the matrix implies that if b1 = b2 then the currents are zero.
b
Another way of seeing this is to notice that if b1 = b2 then 1 is orthogonal to the range
b2
of A − B T C −1 B by (ii) and hence in the nullspace N (A − B T C −1 B).
The number
R=
1
γ
is the ratio of the applied voltage to the resulting current, is the effective resistance of the
network between the two nodes.
So in our example circuit, the effective resistance between nodes 1 and 2 is 5/8.
If the battery voltages are b1 = 0 and b2 = b then the voltages at the remaining nodes are
v3
0
4/5
= −C −1 B
=
b
v4
b
3/5
II.3.11 Two resistors in series
Let’s do a trivial example where we know the answer. If we connect two resistors in series,
the resistances add, and the effective resistance is R1 + R2 . The graph for this example looks
like
83
II Subspaces, Bases and Dimension
1
R1
2
R2
3
The Laplacian for this circuit is
γ1
−γ1
0
L = −γ1 γ1 + γ2 −γ2
0
−γ2
γ2
with γi = 1/Ri , as always. We want the effective resistance between nodes 1 and 3. Although
it is not strictly necessary, it is easier to see what the submatrices A, B and C are if we reorder
the vertices so that the ones we are connecting, namely 1 and 3, come first. This reshuffles
the rows and columns of L yielding
1
3
2
1 γ1
0
−γ1
3 0
γ2
−γ2
2 −γ1 −γ2 γ1 + γ2
Here we have labelled the re-ordered rows and columns with the nodes they represent. Now
the desired submatrices are
γ1 0
C = γ1 + γ2
A=
B = −γ1 −γ2
0 γ2
and
T
A−B C
−1
2
γ1 γ2
1
γ1 γ1 γ2
1 −1
γ1 0
=
B=
−
1
0 γ2
γ1 + γ2 γ1 γ2 γ22
γ1 + γ2 −1
This gives an effective resistance of
R=
as expected.
84
1
1
γ1 + γ2
=
+
= R1 + R2
γ1 γ2
γ1 γ2
II.3 Graphs and Networks
II.3.12 Example: a resistor cube
Hook up resistors along the edges of a cube. If each resistor has resistance Ri = 1, what is
the effective resistance between opposite corners of the cube?
8
7
3
4
5
1
6
2
We will use MATLAB/Octave to solve this problem. To begin we define the Laplace
matrix L. Since each node has three edges connecting it, and all the resistances are 1, the
diagonal entries are all 3. The off-diagonal entries are −1 or 0, depending on whether the
corresponding nodes are connected or not.
>L=[3 -1 0 -1
0 -1 3 -1 0 0
-1 0 0 0 3 -1
0 0 -1 0 0 -1
-1 0 0 0;-1 3 -1
-1 0;-1 0 -1 3 0
0 -1;0 -1 0 0 -1
3 -1;0 0 0 -1 -1
0
0
3
0
0 -1 0 0;
0 -1;
-1 0;
-1 3];
We want to find the effective resistance between 1 and 7. To compute the submatrices A, B
and C it is convenient to re-order the nodes so that 1 and 7 come first. In MATLAB/Octave,
this can be achieved with the following statement.
>L=L([1,7,2:6,8],[1,7,2:6,8]);
In this statement the entries in the first bracket [1,7,2:6,8] indicates the new ordering
of the rows. Here 2:6 stands for 2,3,4,5,6. The second bracket indicates the re-ordering
of the columns, which is the same as for the rows in our case.
Now it is easy to extract the submatrices A, B and C and compute the voltage-to-current
map DN
>N = length(L);
>A = L(1:2,1:2);
>B = L(3:N,1:2);
>C = L(3:N,3:N);
>DN = A - B’*C^(-1)*B;
85
II Subspaces, Bases and Dimension
The effective resistance is the reciprocal of the first entry in DN. The command format rat
gives the answer in rational form. (Note: this is just a rational approximation to the floating
point answer, not an exact rational arithmetic as in Maple or Mathematica.)
>format rat
>R = 1/DN(1,1)
R = 5/6
86
Chapter III
Orthogonality
87
III Orthogonality
III.1 Projections
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the definition of an orthogonal projection matrix
• use propeties of a projection matrix to deduce facts like the orthogonality of the null
space and range.
• compute the orthogonal projection matrix whose range in the span of a given collection
of vectors.
• use orthogonal projection matrices to decompose a vector in to components parallel to
and perpendicular to a given subspace.
• use least squares to compute approximate solutions to systems of equations with no
solutions.
• perform least squares calculations in applications where overdetermined systems arise.
III.1.1 Projections onto lines and planes in R3
Recall the projection of a vector x onto a line containing the non-zero vector a is given by
p = P x, where P is the projection matrix
P =
1
aaT .
kak2
Let’s review why this formula is true, using properties of the dot product. Here is a diagram
of the situation.
a
p
88
θ
x
III.1 Projections
The length of the projected vector p is
kpk = kxk cos(θ) =
a·x
aT x
kakkxk cos(θ)
=
=
kak
kak
kak
To get the vector p start with the unit vector a/kak and stretch it by an amount kpk. This
gives
1
a
=
aaT x
p = kpk
kak
kak2
This can be written p = P x where P is the projection matrix above.
Notice that the matrix P satisfies P 2 = P since
P2 =
1
1
1
aaT aaT =
akak2 aT
aaT = P
4
4
kak
kak
kak2
In addition, P T = P since
PT =
1
1
1
(aaT )T =
(aT )T AT =
aaT = P
2
2
kak
kak
kak2
We will discuss the significance of the two properties P 2 = P and P T = P below.
1
1
2? Let’s calculate
Example: What is the projection of x = 1 in the direction of a =
1
−1
2
the projection matrix P and compute P x and verify that P = P and P T = P .
>x = [1 1 1]’;
>a = [1 2 -1]’;
>P = (a’*a)^(-1)*a*a’
P =
0.16667
0.33333
-0.16667
0.33333
0.66667
-0.33333
-0.16667
-0.33333
0.16667
>P*x
ans =
0.33333
0.66667
-0.33333
>P*P
89
III Orthogonality
ans =
0.16667
0.33333
-0.16667
0.33333
0.66667
-0.33333
-0.16667
-0.33333
0.16667
The projection of x on to the plane orthogonal to a is given by q = x − p.
a
x
p
q
Thus we can write
q = x − P x = (I − P )x = Qx
where
Q=I −P
Notice that, like the matrix P , the matrix Q also satisfies Q2 = Q and QT = Q since
Q2 = (I − P )(I − P ) = I − 2P + P 2 = I − P = Q
and
QT = I T − P T = I − P = Q.
Continuing with the example above, if we want to compute the projection matrix onto the
plane perpendicular to a we compute Q = I − P . Then Qx is the projection of x onto the
plane. We can also check that Q2 = Q.
> Q = eye(3) - P
Q =
0.83333
-0.33333
0.16667
90
-0.33333
0.33333
0.33333
0.16667
0.33333
0.83333
III.1 Projections
>Q*x
ans =
0.66667
0.33333
1.33333
>Q^2
ans =
0.83333
-0.33333
0.16667
-0.33333
0.33333
0.33333
0.16667
0.33333
0.83333
III.1.2 Orthogonal Projections
Any matrix P satisfying P 2 = P is called a projection matrix. If, in addition P T = P then
P is P is called an orthogonal projection. (Warning: this is a different concept than that of
an orthogonal matrix which we will see later.)
The property P 2 = P says that any vector in the range of P is not changed by P , since
P (P x) = P 2 x = P x.
The property P T = P implies that N (P ) = R(P )⊥ . This follows from the orthogonality
relation N (P ) = R(P T )⊥
If P is an orthgonal projection, so is Q = I − P , as you can check. Clearly P + Q = I.
Also
P Q = P (I − P ) = P − P 2 = P − P = 0
and similarly QP = 0. These formulas show that R(P ) = N (Q). To see this, first notice
that if x ∈ R(P ), so that x = P x, then Qx = QP x = 0, which means x ∈ N (Q). Conversely
if x ∈ N (Q) then x = (P + Q)x = P x ∈ R(P ).
As a consequence (since N (Q) = R(QT )⊥ = R(Q)⊥ ) we see that the ranges of P and Q
are orthogonal complements, that is, R(P ) = R(Q)⊥ .
In the example of the last section R(P ) = N (Q) is the line through a while N (P ) = R(Q)
is the plane orthogonal to a.
The orthgonality of P a and Qb implies that
kP a + Qbk2 = (P a + Qb) · (P a + Qb) = kP ak2 + kQbk2
91
III Orthogonality
since the cross terms vanish.
Let P be an orthogonal projection. Let’s show that given any vector x, the vector P x is
the vector in R(P ) that is closest to x. First we compute the square of the distance from P x
to x. This is given by
kP x − xk2 = kQxk2
Now let P y be any other vector in the range R(P ). Then the square of the distance from
P y to x is
kP y − xk2 = kP y − (P + Q)xk2 = kP (y − x) + Qxk2 = kP (y − x)k2 + kQxk2
This implies that kP y − xk2 ≥ kQxk2 = kP x − xk2 , or equivalently kP y − xk ≥ kP x − xk,
with equality exactly when P x = P y.
III.1.3 Least squares solutions
We now consider linear equations
Ax = b
That do not have a solution. This is the same as saying that b 6∈ R(A) What vector x is
closest to being a solution?
b
Ax
Ax−b
R(A) = possible values of Ax
We want to determine x so that Ax is as close as possible to b. In other words, we want
to minimize kAx − bk. This will happen when Ax is the projection of b onto R(A), that
is, Ax = P b, where P is the projection matrix. In this case Qb = (I − P )b is orthogonal
to R(A). But (I − P )b = b − Ax. Therefore (and this is also clear from the picture), we
see that Ax − b is orthogonal to R(A). But the vectors orthogonal to R(A) are exactly the
vectors in N (AT ). Thus the vector we are looking for will satisfy AT (Ax − b) = 0 or the
equation
AT Ax = AT b
This is the least squares equation, and a solution to this equation is called a least squares
solution.
(Aside: We can also use Calculus to derive the least squares equation. We want to minimize
kAx − bk2 . Computing the gradient and setting it to zero results in the same equations.)
92
III.1 Projections
It turns out that the least squares equation always has a solution. Another way of saying
this is R(AT ) = R(AT A). Instead of checking this, we can verify that the orthogonal complements N (A) and N (AT A) are the same. But this is something we showed before, when
we considered the incidence matrix D for a graph.
If x solves the least squares equation, the vector Ax is the projection of b onto the range
R(A), since Ax is the closest vector to x in the range of A. In the case where AT A is
invertible (this happens when N (A) = N (AT A) = {0}), we can obtain a formula for the
projection. Starting with the least squares equation we multiply by (AT A)−1 to obtain
x = (AT A)−1 AT b
so that
Ax = A(AT A)−1 AT b.
Thus the projection matrix is given by
P = A(AT A)−1 AT
Notice that the formula for the projection onto a line through a is a special case of this, since
then AT A = kak2 .
It is worthwhile pointing out that if we say that the solution of the least squares equation
gives the “best” approximation to a solution, what we really mean is that it minimizes the
distance, or equivalently, its square
X
kAx − bk2 =
((Ax)i − bi )2 .
There are other ways of measuring how far Ax is from b, for example the so-called L1 norm
X
kAx − bk1 =
|(Ax)i − bi |
Minimizing the L1 norm will result in a different “best” solution, that may be preferable
under some circumstances. However, it is much more difficult to find!
III.1.4 Polynomial fit
Suppose we have some data points (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ) and we want to fit a polynomial p(x) = a1 xm−1 + a2 xm−2 + · · · + am−1 x + am through them. This is like the Lagrange
interpolation problem we considered before, except that now we assume that n > m. This
means that in general there will no such polynomial. However we can look for the least
squares solution.
To begin, let’s write down the equations that express the desired equalities p(xi ) = yi for
i = 1, . . . m. These can be written in matrix form
93
III Orthogonality
xm−1
xm−2
1
1
xm−1 xm−2
2
2
.
..
..
.
..
..
.
.
..
..
.
.
m−1
m−2
xn
xn
···
···
···
···
···
···
y1
x1 1
x2 1 a
y2
1
.. .. ..
. . a2 .
.. = ..
.. ..
.
. .
.
..
.. .. am
.
. .
yn
xn 1
or Aa = y. where A is a submatrix of the Vandermonde matrix. To find the least squares
approximation we solve AT Aa = AT y. In a homework problem, you are asked to do this
using MATLAB/Octave.
In case where the polynomial has degree one this is a straight line fit, and the equation we
want to solve are
x1 1
y1
x2 1 y2
a1
= .
..
a
.
2
..
xn 1
yn
These equations will not have a solution (unless the points really do happen to lie on the
same line.) To find the least squares solution, we compute
x1 1
P 2 P x1 x2 · · · xn x2 1
xi
x
= P i
1 1 · · · 1 ...
x
n
i
xn 1
and
x1 x2 · · ·
1 1 ···
y1
P
xi y i
xn y2
P
=
yi
1 ...
yn
This results in the familiar equations
P 2 P P
a
x
x
y
x
1
i
i
i
i
P
= P
xi
n
yi
a2
which are easily solved.
III.1.5 Football rankings
We can try to use least squares to rank football teams. To start with, suppose we have three
teams. We pretend each team has a value v1 , v2 and v3 such that when two teams play, the
94
III.1 Projections
difference in scores is the difference in values. So, if the season’s games had the following
results
1 vs. 2 30 40
1 vs. 2 20 40
2 vs. 3 10 0
3 vs. 1 0 5
3 vs. 2 5 5
then the vi ’s would satisfy the equations
v2 − v1 = 10
v2 − v1 = 20
v2 − v3 = 10
v1 − v3 = 5
v2 − v3 = 0
Of course, there is no solution to these equations. Nevertheless we can find the least squares
solution. The matrix form of the equations is
Dv = b
with
The least squares equation is
−1
−1
D=
0
−1
0
1
0
1
0
1 −1
0
1
1 −1
10
20
b=
10
5
0
D T Dv = D T v
or
3 −2 −1
−35
−2 4 −2 v = 40
−1 −2 3
−5
Before going on, notice that D is an incidence matrix. What is the graph? (Answer: the
nodes are the teams and they are joined by an edge with the arrow pointing from the losing
team to the winning team. This graph may have more than one edge joining to nodes, if
two teams play more than once. This is sometimes called a multi-graph.). We saw that in
this situation N (D) is not empty, but contains vectors whose entries are all the same. The
situation is the same as for resistances, it is only differences in vi ’s that have a meaning.
We can solve this equation in MATLAB/Octave. The straightforward way is to compute
>L = [3 -2 -1;-2 4 -2;-1 -2 3];
95
III Orthogonality
>b = [-35; 40; -5];
>rref([L b])
ans =
1.00000
0.00000
0.00000
0.00000
1.00000
0.00000
-1.00000
-1.00000
0.00000
-7.50000
6.25000
0.00000
As expected, the solution is not unique. The general solution, depending on the parameter
s is
−7.5
1
v = s 1 + 6.25
1
0
We can choose s so that the vi for one of the teams is zero. This is like grounding
a node
0
in a circuit. So, by choosing s = 7.5, s = −6.25 and s = 0 we obtain the solutions 13.75,
7.5
−13.75
−7.5
0 or 6.25 .
−6.25
0
Actually,
it is easier to compute a solution with one of the vi ’s equal to zero directly. If
0
v
v = v2 then v2 = 2 satisfies the equation L2 v2 = b2 where the matrix L2 is the bottom
v3
v3
right 2 × 2 block of L and b2 are the last two entries of b.
>L2 = L(2:3,2:3);
>b2 = b(2:3);
>L2\b2
ans =
13.7500
7.5000
We can try this on real data. The football scores for the 2007 CFL season can be found at
http://www.cfl.ca/index.php?module=sked&func=view&year=2007. The differences in
scores for the first 20 games are in cfl.m. The order of the teams is BC, Calgary, Edmonton,
Hamilton, Montreal, Saskatchewan, Toronto, Winnipeg. Repeating the computation above
for this data we find the ranking to be (running the file cfl.m)
96
III.1 Projections
v =
0.00000
-12.85980
-17.71983
-22.01884
-11.37097
-1.21812
0.87588
-20.36966
Not very impressive, if you consider that the second-lowest ranked team (Winnipeg) ended
up in the Grey Cup game!
97
III Orthogonality
III.2 Orthonormal bases and Orthogonal Matrices
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the definition of an orthonormal basis.
• compute the coefficients in the expansion of an orthonormal basis
• compute the norm of a vector from its coefficients in an orthonormal basis
• write down the definition of an orthogonal matrix
• recognize an orthogonal matrix by its rows or columns
• know how to characterize an orthogonal matrix by its action on vectors
III.2.1 Orthonormal bases
A basis q1 , q2 , . . . is called orthonormal if
1. kqi k = 1 for every i (normal)
2. qi · qj = 0 for i 6= j (ortho).
The standard basis for Rn given by
1
0
0
1
e1 = 0 , e2 = 0 ,
..
..
.
.
0
0
e3 = 1 ,
..
.
···
is an orthonormal basis for Rn . Another orthonormal basis for R2 is
1 −1
1 1
, q1 = √
q1 = √
1
2 1
2
If you expand a vector in an orthonormal basis, it’s very easy to find the coefficients in the
expansion. Suppose
v = c1 q1 + c2 q2 + · · · + cn qn
98
III.2 Orthonormal bases and Orthogonal Matrices
for some orthonormal basis q1 , q2 , . . .. Then, if we take the dot product of both sides with
qk , we get
qk · v = c1 qk · q1 + c2 qk · q2 + · · · + ck qk · qk · · · + cn qk · qn
= 0 + 0 + · · · + ck + · · · + 0
= ck
This gives a convenient formula for each ck . For example, in the expansion
1 −1
1 1
1
+ c2 √
= c1 √
1
2
2 1
2
we have
1 1
3
1
c1 = √
·
=√
2
2 1
2
1 −1
1
1
c2 = √
·
=√
1
2
2
2
Notice also that the norm of v is easily expressed in terms of the coefficients ci . We have
kvk2 = v · v
= (c1 q1 + c2 q2 + · · · + cn qn ) · (c1 q1 + c2 q2 + · · · + cn qn )
= c21 + c22 + · · · + c2n
Another way of saying this is that the vector c = [c1 , c2 , . . . , cn ] of coefficients has the same
norm as v.
III.2.2 Orthogonal matrices
An n × n matrix Q is called orthogonal if QT Q = I (equivalently if QT = Q−1 ).
columns of Q are q1 , q2 , . . . , qn then Q is orthogonal if
T
q1 · q1 q1 · q2 · · · q1 · qn
1 0 ···
q1
q2 · q1 q2 · q2 · · · q2 · qn 0 1 · · ·
qT
2
QT Q = . q1 q2 · · · qn = .
..
..
.. = .. ..
..
..
..
.
.
. . .
.
qTn
qn · q1 qn · q2 · · ·
qn · qn
0 0 ···
This is the same as saying that the columns of Q form an orthonormal basis.
If the
0
0
.. .
.
1
Another way of recognizing orthogonal matrices is by their action on vectors. Suppose Q
is orthogonal. Then
kQvk2 = (Qv) · (Qv) = v · (QT Qv) = v · v = kvk2
99
III Orthogonality
This implies that kQvk = kvk. In other words, orthogonal matrices don’t change the lengths
of vectors.
The converse is also true. If a matrix Q doesn’t change the lengths of vectors then it must
be orthogonal. To see this, suppose that kQvk = kvk for every v. Then the calculation
above shows that v · (QT Qv) = v · v for every v. Applying this to v + w we find
(v + w) · QT Q(v + w) = (v + w) · (v + w)
Expanding, this gives
v · (QT Qv) + w · (QT Qw) + v · (QT Qw) + w · (QT Qv) = v · v + w · w + v · w + w · v
Since v · (QT Qv) = v · v and w · (QT Qw) = w · w we can cancel these terms. Also
w · (QT Qv) = ((QT Q)T w) · v = (QT Qw) · v = v · (QT Qw). So on each side of the equation,
the two remaining terms are the same. Thus
v · (QT Qw) = v · w
This equation holds for every choice of vectors v and w. If v = ei and w = ej then the left
side is the i, jth matrix element Qi,j of Q while the right side is the ei · ej , which is i, jth
matrix element of the identity matrix. Thus QT Q = I and Q is orthogonal.
We can recast the problem of finding coefficients c1 , c2 , . . . , cn in the expansion v = c1 q1 +
c2 q2 + · · · + cn qn in an orthonormal basis as the solution of the matrix equation Qc = v
where Q is the orthogonal matrix whose columns contain the orthonormal basis vectors. The
solution is obtained by multiplying by QT . Since QT Q = I multiplying both sides by QT
gives c = QT v. The fact that kck = kvk follows from the length preserving property of
orthogonal matrices.
Recall that for square matrices a left inverse is automatically also a right inverse. So if
= I then QQT = I too. This means that QT is an orthogonal matrix whenever Q
is. This proves the (non-obvious) fact that if the columns of an square matrix form an
orthonormal basis, then so do the rows!
QT Q
A set G of invertible matrices is called a (matrix) group if
1. I ∈ G (G contains the identity matrix)
2. If A, B ∈ G then AB ∈ G (G is closed under matrix multiplication)
3. If A ∈ G then A is invertible and A−1 ∈ G (G is closed under taking the inverse)
In a homework problem, you are asked to verify that the set of n × n orthogonal matrices
is a group.
100
III.3 Complex vector spaces and inner product
III.3 Complex vector spaces and inner product
Prerequisites and Learning Goals
From your work in previous courses, you should be able to
• perform arithmetic with complex numbers.
• write down the definition of complex conjugate, modulus and argument of a complex
number
• write down the definition of the complex exponential, addition formula, differentiation
and integration of complex exponentials.
After completing this section, you should be able to
• write down the definition of complex inner product and the norm of a complex vector
• write down the definition of the matrix adjoint, its relation to the complex inner product.
• write down the definition of a unitary matrix and its properties.
• use complex numbers in MATLAB/Octave computations, specifically real(z), imag(z),
conj(z), abs(z), exp(z) and A’ for complex matrices.
III.3.1 Review of complex numbers
x
Complex numbers can be thought of as points on the (x, y) plane. The point
, thought
y
of as a complex number, is written x + iy (or x + jy if you are an electrical engineer).
If z = x + iy then x is called the real part of z and y is called the imaginary part of z.
Complex numbers are added just as if they were vectors in two dimensions. If z = x + iy
and w = s + it, then
z + w = (x + iy) + (s + it) = (x + s) + i(y + t)
To multiply two complex numbers, just remember that i2 = −1. So if z = x + iy and
w = s + it, then
zw = (x + iy)(s + it) = xs + i2 yt + iys + ixt = (xs − yt) + i(xt + ys)
101
III Orthogonality
The modulus of a complex number, denoted |z| is simply the length of the corresponding
vector in two dimensions. If z = x + iy
p
|z| = |x + iy| = x2 + y 2
An important property is
|zw| = |z||w|
The complex conjugate of a complex number z, denoted z̄, is the reflection of z across the
x axis. Thus x + iy = x − iy. Thus complex conjugate is obtained by changing all the i’s to
−i’s. We have
zw = z̄ w̄
and
zz̄ = |z|2
This last equality is useful for simplifying fractions of complex numbers by turning the
denominator into a real number, since
z w̄
z
=
w
|w|2
For example, to simplify (1 + i)/(1 − i) we can write
1+i
(1 + i)2
1 − 1 + 2i
=
=
=i
1−i
(1 − i)(1 + i)
2
A complex number z is real (i.e. the y part in x + iy is zero) whenever z̄ = z. We also have
the following formulas for the real and imaginary part. If z = x + iy then x = (z + z̄)/2 and
y = (z − z̄)/(2i)
We define the exponential, eit , of a purely imaginary number it to be the number
eit = cos(t) + i sin(t)
lying on the unit circle in the complex plane.
The complex exponential satisfies the familiar rule ei(s+t) = eis eit since by the addition
formulas for sine and cosine
ei(s+t) = cos(s + t) + i sin(s + t)
= cos(s) cos(t) − sin(s) sin(t) + i(sin(s) cos(t) + cos(s) sin(t))
= (cos(s) + i sin(s))(cos(t) + i sin(t))
= eis eit
The exponential of a number that has both a real and imaginary part is defined in the
natural way.
ea+ib = ea eib = ea (cos(b) + i sin(b))
102
III.3 Complex vector spaces and inner product
The derivative of a complex exponential is given by the formula
d (a+ib)t
e
= (a + ib)e(a+ib)t
dt
while the anti-derivative, for (a + ib) 6= 0 is
Z
e(a+ib)t dt =
1
e(a+ib)t + C
(a + ib)
If (a + ib) = 0 then e(a+ib)t = e0 = 1 so in this case
Z
Z
(a+ib)t
e
dt = dt = t + C
III.3.2 Complex vector spaces and inner product
So far in this course, our scalars have been real numbers. We now want to allow complex
numbers. The basic example of a complex vector space is the space Cn of n-tuples of complex
numbers. Vector addition and scalar multiplication are defined as before:
sz1
z1
z1 + w1
w1
z1
z2 sz2
z2 w2 z2 + w2
, s .. = .. ,
.. + .. =
..
. .
. .
.
zn
wn
zn
zn + wn
szn
where now zi , wi and s are complex numbers.
For complex matrices (or vectors) we define the complex conjugate matrix (or vector) by
conjugating each entry. Thus, if A = [ai,j ], then
A = [ai,j ].
The product rule for complex conjugation extends to matrices and we have
AB = ĀB̄
w1
z1
w2
z2
The inner product of two complex vectors w = . and z = . is defined by
..
..
wn
T
hw, zi = w z =
n
X
zn
w i zi
i=1
103
III Orthogonality
With this definition the norm of z is always positive since
2
hz, zi = kzk =
n
X
i=1
|zi |2
For complex matrices and vectors we have to modify the rule for bringing a matrix to the
other side of an inner product.
hw, Azi = wT Az
= (AT w)T z
T
T
= (A w)
z
T
= hA w, zi
This leads to the definition of the adjoint of a matrix
T
A∗ = A .
(In physics you will also see the notation A† .) With this notation hw, Azi = hA∗ w, zi.
The complex analogue of an orthogonal matrix is called a unitary matrix. A unitary matrix
U is a square matrix satisfying
U ∗ U = U U ∗ = I.
Notice that a unitary matrix with real entries is an orthogonal matrix since in that case
U ∗ = U T . The columns of a unitary matrix form an orthonormal basis (with respect to the
complex inner product.)
MATLAB/Octave deals seamlessly with complex matrices and vectors. Complex numbers
can be entered like this
>z= 1 + 2i
z =
1 + 2i
There is a slight danger here in that if i has be defined to be something else (e.g. i =16)
then z=i would set z to be 16. You could use z=1i to get the desired result, or use the
alternative syntax
>z= complex(0,1)
z =
104
0 + 1i
III.3 Complex vector spaces and inner product
The functions real(z), imag(z), conj(z), abs(z) compute the real part, imaginary part,
conjugate and modulus of z.
The function exp(z) computes the complex exponential if z is complex.
If a matrix A has complex entries then A’ is not the transpose, but the adjoint (conjugate
transpose).
>z = [1; 1i]
z =
1 + 0i
0 + 1i
z’
ans =
1 - 0i
0 - 1i
Thus the square of the norm of a complex vector is given by
>z’*z
ans =
2
This gives the same answer as
>norm(z)^2
ans =
2.0000
(Warning: the function dot in Octave does not compute the correct inner product for complex vectors (it doesn’t take the complex conjugate). This is probably a bug. In MATLAB
dot works correctly for complex vectors.)
105
III Orthogonality
III.3.3 Vector spaces of complex-valued functions
Let [a, b] be an interval on the real line. Recall that we introduced the vector space of real
valued functions defined for x ∈ [a, b]. The vector sum f + g of two functions f and g was
defined to be the function you get by adding the values, that is, (f + g)(x) = f (x) + g(x)
and the scalar multiple sf was defined similarly by (sf )(x) = sf (x).
In exactly the same way, we can introduce a vector space of complex valued functions.
The independent variable x is still real, taking values in [a, b]. But now the values f (x) of
the functions may be complex. Examples of complex valued functions are f (x) = x + ix2 or
f (x) = eix = cos(x) + i sin(x).
Now we introduce the inner product of two complex valued functions on [a, b]. In analogy
with the inner product for complex vectors we define
Z
hf, gi =
b
f (x)g(x)dx
a
and the assoicated norm defined by
2
kf k = hf, f i =
Z
b
a
|f (x)|2 dx
For real valued functions we can ignore the complex conjugate.
Example: the inner product of f (x) = 1 + ix and g(x) = x2 over the interval [0, 1] is
2
h1 + ix, x i =
Z
1
0
2
(1 + ix) · x dx =
Z
0
1
2
(1 − ix) · x dx =
Z
1
0
x2 − ix3 dx =
1
1
−i
3
4
It will often happen that a function, like f (x) = x is defined for all real values of x. In this
case we can consider inner products and norms for any interval [a, b] including semi-infinite
and infinite intervals, where a may be −∞ or b may be +∞. Of course the values of the
inner product an norm depend on the choice of interval.
There are technical complications when dealing with spaces of functions. In this course we
will deal with aspects of the subject where these complications don’t play an important role.
However, it is good to aware that they exist, so we will mention a few.
One complication is that the integral defining the inner product may not exist. For example
for the interval (−∞, ∞) = R the norm of f (x) = x is infinite since
Z ∞
|x|2 dx = ∞
−∞
Even if the interval is finite, like [0, 1], the function might have a spike. For example, if
f (x) = 1/x then
Z 1
1
dx = ∞
2
|x|
0
106
III.3 Complex vector spaces and inner product
too. To overcome this complication we agree to restrict our attention to square integrable
functions. For any interval [a, b], these are the functions f (x) for which |f (x)|2 is integrable.
They form a vector space that is usually denoted L2 ([a, b]). It is an example of a Hilbert space
and is important in Quantum Mechanics. The L in this notation indicates that the integrals
should be defined as Lebesgue integrals rather than as Riemann integrals usually taught in
elementary calculus courses. This plays a role when discussing convergence theorems. But
for any functions that come up in this course, the Lebesgue integral and the Riemann integral
will be the same.
The question of convergence is another complication that arises in infinite dimensional
vector spaces of functions. When discussing infinite orthonormal bases, infinite linear combinations of vectors (functions) will appear. There are several possible meanings for an
equation like
∞
X
ci φi (x) = φ(x).
i=0
since we are talking about convergence of an infinite series of functions. The most obvious
interpretation is that for every fixed value of x the infinite sum of numbers on the left hand
side equals the number on the right.
P
Here is another interpretation: the difference of φ and the partial sums N
i=0 ci φi tends to
2
zero when measured in the L norm, that is
lim k
N →∞
N
X
i=0
ci φi − φk = 0
With this definition, it might happen that there are individual values of x where the first
equation doesn’t hold. This is the meaning that we will give to the equation.
107
III Orthogonality
III.4 Fourier series
Prerequisites and Learning Goals
After completing this section, you should be able to
• compute the Fourier series (in real and complex form) of a function defined on the
interval [0, L]
• interpret each of these series as the expansion of a function (vector) in an infinite
orthogonal basis.
• use Parseval’s formula to sum certain infinite series
• use MATLAB/Octave to plot the partial sums of Fourier series.
• explain what an amplitude-frequency plot is and be able to compute it in examples.
III.4.1 An infinite orthonormal basis for L2 ([a, b])
Let [a, b] be an interval of length L = b − a. For every integer n, define the function
en (x) = e2πinx/L .
Then infinite collection of functions
{. . . , e−2 , e−1 , e0 , e1 e2 , . . .}
forms
an orthonormal basis for the space L2 ([a, b]), except that each function en has norm
√
L instead of 1. (Since this is the usual normalization,
√ we will stick with it. To get a true
orthonormal basis, we must divide each function by L.)
√
Let’s verify that these function form an orthonormal set (scaled by L). To compute the
norm we calculate
Z b
e2πinx/L e2πinx/L dx
ken k2 = hen , en i =
a
Z b
e−2πinx/L e2πinx/L dx
=
a
Z b
=
1dx.
a
=L
108
III.4 Fourier series
This shows that ken k =
orthogonal.
√
L for every n. Next we check that if n 6= m then en and em are
hen , em i =
=
Z
Z
b
e−2πinx/L e2πimx/L dx
a
b
e2πi(m−n)x/L dx
a
b
L
e2πi(m−n)x/L 2πi(m − n)
x=a
L
e2πi(m−n)b/L − e2πi(m−n)a/L
=
2πi(m − n)
=0
=
Here we used that e2πi(m−n)b/L = e2πi(m−n)(b−a+a)/L = e2πi(m−n) e2πi(m−n)a/L = e2πi(m−n)a/L .
This
√ shows that the functions {. . . , e−2 , e−1 , e0 , e1 e2 , . . .} form an orthonormal set (scaled by
L).
To show these functions form a basis we have to verify that they span the space L2 [a, b].
In other words, we must show that any function f ∈ L2 [a, b] can be written as an infinite
linear combination
∞
∞
X
X
cn e2πinx/L .
cn en (x) =
f (x) =
n=−∞
n=−∞
This is a bit tricky, since it involves infinite series of functions. For a finite dimensional space,
to show that an orthogonal set forms a basis, it suffices to count that there are the same
number of elements in an orthogonal set as there are dimensions in the space. For an infinite
dimensional space this is no longer true. For example, the set of en ’s with n even is also an
infinite orthonormal set, but it doesn’t span all of L2 [a, b].
In this course, we will simply accept that it is true that {. . . , e−2 , e−1 , e0 , e1 e2 , . . .} span
L2 [a, b]. Once we accept this fact, it is very easy to compute the coefficients in a Fourier
expansion. The procedure is the same as in finite dimensions. Starting with
∞
X
f (x) =
cn en (x)
n=−∞
we simply take the inner product of both sides with em . The only term in the infinite sum
that survives is the one with n = m. Thus
hem , f i =
and we obtain the formula
cm =
1
L
∞
X
n=−∞
Z
b
cn hem , en i = cm L
e−2πimx/L f (x)dx
a
109
III Orthogonality
III.4.2 Real form of the Fourier series
Fourier series are often written in terms of sines and cosines as
∞
a0 X
f (x) =
(an cos(2πnx/L) + bn sin(2πnx/L))
+
2
n=1
To obtain this form, recall that
e±2πinx/L = cos(2πnx/L) ± i sin(2πnx/L)
Using this formula we find
∞
X
2πnx/L
cn e
= c0 +
n=−∞
= c0 +
= c0 +
∞
X
n=1
∞
X
n=1
∞
X
n=1
2πnx/L
cn e
+
∞
X
n=1
c−n e−2πnx/L
cn (cos(2πnx/L) + i sin(2πnx/L)) +
∞
X
n=1
c−n (cos(2πnx/L) − i sin(2πnx/L))
((cn + c−n ) cos(2πnx/L) + i(cn − c−n ) sin(2πnx/L)))
Thus the real form of the Fourier series holds with
a0 = 2c0
an = cn + c−n
bn = icn − ic−n
for
n>0
for n > 0.
Equivalently
a0
2
an bn
+
for n > 0
cn =
2
2i
a−n b−n
−
for n < 0.
cn =
2
2i
c0 =
The coefficients an and bn in the real form of the Fourier series can also be obtained directly.
The set of functions
{1/2, cos(2πx/L), cos(4πx/L), cos(6πx/L), . . . , sin(2πx/L), sin(4πx/L), sin(6πx/L), . . .}
p
also form an orthogonal basis where each vector has norm L/2. This leads to the formulas
2
an =
L
110
Z
b
cos(2πnx/L)f (x)
a
III.4 Fourier series
for n = 0, 1, 2, . . . and
2
bn =
L
Z
b
sin(2πnx/L)f (x)
a
for n = 1, 2, . . .. The desire to have the formula for an work out for n = 0 is the reason for
dividing by 2 in the constant term a0 /2 in the real form of the Fourier series.
One advantage of the real form of the Fourier series is that if f (x) is a real valued function,
then the coefficients an and bn are real too, and the Fourier series doesn’t involve any complex
numbers. However, it is often to calculate the coefficients cn because exponentials are easier
to integrate than sines and cosines.
III.4.3 An example
Let’s compute the Fourier coefficients for the square wave function. In this example L = 1.
1 if 0 ≤ x ≤ 1/2
f (x) =
−1 if 1/2 < x ≤ 1
If n = 0 then e−i2πnx = e0 = 1 so c0 is simply the integral of f .
c0 =
Z
1
f (x)dx =
0
Z
1/2
0
1dx −
Z
1
1dx = 0
1/2
Otherwise, we have
cn =
=
Z
Z
1
e−i2πnx f (x)dx
0
1/2
−i2πnx
e
0
dx −
Z
1
e−i2πnx dx
1/2
e−i2πnx x=1/2 e−i2πnx x=1
=
−
−i2πn x=0
−i2πn x=1/2
2 − 2eiπn
=
2πin
0
if n is even
=
2/iπn if n is odd
Thus we conclude that
f (x) =
∞
X
n=−∞
n odd
2 i2πnx
e
iπn
To see how well this series is approximating f (x) we go back to the real form of the series.
Using an = cn + c−n and bn = icn − ic−n we find that an = 0 for all n, bn = 0 for n even and
111
III Orthogonality
bn = 4/πn for n odd. Thus
f (x) =
∞
∞
X
X
4
4
sin(2πnx) =
sin(2π(2n + 1)x)
πn
π(2n + 1)
n=1
n=0
n odd
We can use MATLAB/Octave to see how well this series is converging. The file ftdemo1.m
contains a function that take an integer N as an argument and plots the sum of the first
2N + 1 terms in the Fourier series above. Here is a listing:
function ftdemo1(N)
X=linspace(0,1,1000);
F=zeros(1,1000);
for n=[0:N]
F = F + 4*sin(2*pi*(2*n+1)*X)/(pi*(2*n+1));
end
plot(X,F)
end
Here are the outputs for N = 0, 1, 2, 10, 50:
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-0.2
0.2
0.4
0.6
0.8
1
1.2
-1.5
-0.2
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-0.2
112
0
1.5
0
0.2
0.4
0.6
0.8
1
1.2
-1.5
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0
0.2
0.4
0.6
0.8
1
1.2
III.4 Fourier series
1.5
1
0.5
0
-0.5
-1
-1.5
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
III.4.4 Parseval’s formula
If v1 , v2 , . . . , vn is an orthonormal basis in a finite dimensional vector space and the vector
v has the expansion
n
X
ci vi
v = c1 v1 + · · · + cn vn =
i=1
then, taking the inner product of v with itself, and using the fact that the basis is orthonormal, we obtain
n
n X
n
X
X
|ci |2
ci cj hvi , vj i =
hv, vi =
i=1
i=1 j=1
The same formula is true in Hilbert space. If
∞
X
f (x) =
cn en (x)
n=−∞
Then
Z
0
1
|f (x)|2 dx = hf, f i =
In the example above, we have hf, f i =
1=
n=∞
X
n=−∞
n odd
or
R1
0
∞
X
n=−∞
|cn |2
1dx = 1 so we obtain
n=∞
∞
X 4
1
8 X
4
=
2
=
2
2
2
2
2
π n
π n
π n=0 (2n + 1)2
n=0
n odd
∞
X
1
π2
=
(2n + 1)2
8
n=0
III.4.5 Interpretation of Fourier series
What is the meaning of a Fourier series in a practical example? Consider the sound made by
a musical instrument in a time interval [0, T ]. This sound can be represented by a function
113
III Orthogonality
y(t) for t ∈ [0, T ], where y(t) is the air pressure at a point in space, for example, at your
eardrum.
A complex exponential e2πiωt = cos(2πωt) ± i sin(2πωt) can be thought of as a pure oscillation with frequency ω. It is a periodic function whose values are repeated when t increases
by ω −1 . If t has units of time (seconds) then ω has units of Hertz (cycles per second). In
other words, in one second the function e2πiωt cycles though its values ω times.
The Fourier basis functions can be written as e2πiωn t with ωn = n/T . Thus Fourier’s
theorem states that for t ∈ [0, T ]
y(t) =
∞
X
cn e2πiωn t .
n=−∞
In other words, the audio signal y(t) can be synthesized as a superposition of pure oscillations
with frequencies ωn = n/T . The coefficients cn describe how much of the frequency ωn is
present in the signal. More precisely, writing the complex number cn as cn = |cn |e2πiτn we
have cn e2πiωn t = |cn |e2πi(ωn t+τn ) . Thus |cn | represents the amplitude of the oscillation with
frequency ωn while τn represents a phase shift.
A frequency-amplitude plot for y(t) is a plot of the points (ωn , |cn |). It should be thought
of as a graph of the amplitude as a function of frequency and gives a visual representation
of how much of each frequency is present in the signal.
If y(t) is defined for all values of t we can use any interval that we want and expand the
restriction of y(t) to this interval. Notice that the frequencies ωn = n/T in the expansion
will be different for different values of T .
Example: Let’s illustrate this with the function y(t) = e2πit and intervals [0, T ]. This
function is itself a pure oscillation with frequency ω = 1. So at first glance one would expect
that there will be only one term in the Fourier expansion. This will turn out to be correct
if number 1 is one of the available frequencies, that is, if there is some value of n for which
ωn = n/T = 1. (This happens if T is an integer.) Otherwise, it is still possible to reconstruct
y(t), but more frequencies will be required. In this case we would expect that |cn | should be
large for ωn close to 1. Let’s do the calculation. Fix T . Let’s first consider the case when T
is an integer. Then
Z
1 T −2πint/T 2πit
e
e dt
cn =
T 0
Z
1 T 2πi(1−n/T )t
e
dt
=
T 0
(
1
n=T
=
1
2πi(T
−n)
0
− e = 0 n 6= T,
2T πi(1−n/T ) e
114
III.4 Fourier series
as expected. Now let’s look at what happens when T is not an integer. Then
1
cn =
T
Z
T
e−2πint/T e2πit dt
0
1
=
e2πi(T −n) − 1
2πi(T − n)
A calculation (that we leave as an exercise) results in
p
2 − 2 cos(2πT (1 − ωn ))
|cn | =
2πT |1 − ωn |
We can use MATLAB/Octave to do an amplitude-frequency plot. Here are the commands
for T = 10.5 and T = 100.5
N=[-200:200];
T=10.5;
omega=N/T;
absc=sqrt(2-2*cos(2*pi*T*(1-omega)))./(2*pi*T*abs(1-omega));
plot(omega,absc)
T=100.5;
omega=N/T;
absc=sqrt(2-2*cos(2*pi*T*(1-omega)))./(2*pi*T*abs(1-omega));
hold on;
plot(omega,absc, ’r’)
Here is the result
0.6
0.5
0.4
0.3
0.2
0.1
0
-20
-15
-10
-5
0
5
10
15
20
As expected, the values of |cn | are largest when ωn is close to 1.
115
III Orthogonality
III.5 The Discrete Fourier Transform
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the definition of the discrete Fourier transform, and compute the matrix
that implements it.
• explain why the Fast Fourier transform algorithm is a faster method.
• compute the discrete Fourier transform of a vector using the fft algorithm.
• compute the fft and its inverse using MATLAB/Octave.
• compute a frequency-amplitude plot for a sampled signal using MATLAB/Octave, and
interpret the result.
III.5.1 Definition
In the previous section we saw that the functions ek (x) = e2πikx for k ∈ Z form an infinite
orthonormal basis for the Hilbert space of functions L2 ([0, 1]). Now we will introduce a
discrete, finite dimensional version of this basis.
To motivate the definition of this basis, imagine taking a function defined on the interval
[0, 1] and sampling it at the point at the N points 0, 1/N, 2/N, . . . , j/N, . . . , (N − 1)/N . If
we do this to the basis functions ek (x) we end up with vectors ek given by
where
ek =
e2πi0k/N
e2πik/N
e2πi2k/N
..
.
e2πi(N −1)k/N
=
1
k
ωN
2k
ωN
..
.
(N −1)k
ωN
ωN = e2πi/N
The complex number ωN lies on the unit circle, that is, |ωN | = 1. Moreover ωN is a primitive
N = 1 and ω j 6= 1 unless j is a multiple of N .
N th root of unity. This means that ωN
N
k+N
k ω N = ω k we see that e
= ωN
Because ωN
k+N = ek . Thus, although the vectors ek are
N
N
defined for every integer k, they start repeating themselves after N steps. Thus there are
only N distinct vectors, e0 , e1 , . . . , eN −1 .
116
III.5 The Discrete Fourier Transform
These vectors, ek for k = 0, . . . , N − 1 form an orthogonal basis for CN . To see this we use
the formula for the sum of a geometric series:
N
−1
N
r=1
X
j
N
r = 1−r
r 6= 1
j=0
1−r
Using this formula, we compute
hek , el i =
N
−1
X
N
−1
X
lj
ωN kj ωN
=
j=0
(l−k)j
ωN
j=0
l=k
N
(l−k)N
= 1 − ωN
=0 l=
6 k
l−k
1 − ωN
Now we can expand any vector f ∈ CN in this basis. Actually, to make our discrete Fourier
transform agree with MATLAB/Octave we divide each basis vector by N . Then we obtain
N −1
1 X
cj ej
f=
N
j=0
where
ck = hek , f i =
N
−1
X
e−2πikj/N fj
j=0
The map that send the vector f to the vector of coefficients c = [c0 , . . . , cN −1 ]T is the
discrete Fourier transform. We can write this in matrix form as
f = F −1 c
c = Ff,
where the matrix F −1 has the vectors ek as its columns. Since this vectors are an orthogonal
basis, the inverse is the transpose, up to a factor of N . Explicitly
1
1
1
F =
.
.
.
1
ωN
ω 2N
..
.
1
ω 2N
ω 4N
..
.
···
···
···
2(N −1)
N −1
1 ωN
ωN
and
F
−1
1
1
1
1
=
N ..
.
1
ωN
ωN 2
..
.
1
ωN 2
ωN 4
..
.
···
1
N −1
ωN
2(N −1)
ωN
..
.
(N −1)(N −1)
ωN
···
···
···
1 ωN N −1 ωN 2(N −1) · · ·
1
ωN N −1
ωN 2(N −1)
..
.
ωN (N −1)(N −1)
117
III Orthogonality
The matrix F̃ = N −1/2 F is a unitary matrix (F̃ −1 = F̃ ∗ ). Recall that unitary matrices
preserve the length of complex vectors. This implies that the lengths of the vectors f =
[f0 , f1 , . . . , fN −1 ] and c = [c0 , c1 , . . . , cN −1 ] are related by
N kck2 = kf k2
or
N
N
−1
X
k=0
|ck |2 =
N
−1
X
k=0
|fk |2
This is the discrete version of Parseval’s formula.
III.5.2 The Fast Fourier transform
Multiplying an N ×N matrix with a vector of length N normally requires N 2 multiplications,
since each entry of the product requires N , and there are N entries. It turns out that the
discrete Fourier transform, that is, multiplication by the matrix F , can be carried out using
only N log2 (N ) multiplications (at least if N is a power of 2). The algorithm that achieves
this is called the Fast Fourier Transform, or FFT. This represents a tremendous saving in
time: calculations that would require weeks of computer time can be carried out in seconds.
The basic idea of the FFT is to break the sum defining the Fourier coefficients ck into a
sum of the even terms and a sum of the odd terms. Each of these turns out to be (up to
a factor we can compute) a discrete Fourier transform of half the length. This idea is then
applied recursively. Starting with N = 2n and halving the size of the Fourier transform at
each step, it takes n = log2 (N ) steps to arrive at Fourier transforms of length 1. This is
where the log2 (N ) comes in.
To simplify the notation, we will ignore the factor of 1/N in the definition of the discrete
Fourier transform (so one should divide by N at the end of the calculation.) We now also
assume that
N = 2n
so that we can divide N by 2 repeatedly. The basic formula, splitting the sum for ck into a
118
III.5 The Discrete Fourier Transform
sum over odd and even j’s is
ck =
N
−1
X
e−i2πkj/N fj
j=0
=
N
−1
X
e−i2πkj/N fj +
N
−1
X
j=0
j odd
j=0
j even
N/2−1
=
N/2−1
X
e
X
e−i2πkj/(N/2) f2j + e−i2πk/N
−i2πk2j/N
f2j +
j=0
X
e−i2πk(2j+1)/N f2j+1
j=0
N/2−1
=
e−i2πkj/N fj
N/2−1
X
e−i2πkj/(N/2) f2j+1
j=0
j=0
Notice that the two sums on the right are discrete Fourier transforms of length N/2.
To continue, it is useful to write the integers j in base 2. Lets assume that N = 23 = 8.
Once you understand this case, the general case N = 2n will be easy. Recall that
0 = 000
(base 2)
1 = 001
(base 2)
2 = 010
(base 2)
3 = 011
(base 2)
4 = 100
(base 2)
5 = 101
(base 2)
6 = 110
(base 2)
7 = 111
(base 2)
The even j’s are the ones whose binary expansions have the form ∗ ∗ 0, while the odd j’s
have binary expansions of the form ∗ ∗ 1.
For any pattern of bits like ∗ ∗ 0, I will use the notation F <pattern> to denote the discrete
Fourier transform where the input data is given by all the fj ’s whose j’s have binary expansion
fitting the pattern. Here are some examples. To start, Fk∗∗∗ = ck is the original discrete
Fourier transform, since every j fits the pattern ∗ ∗ ∗. In this example k ranges over 0, . . . , 7,
that is, the values start repeating after that.
Only even j’s fit the pattern ∗ ∗ 0, so F ∗∗0 is the discrete Fourier transform of the even j’s
given by
N/2−1
X
Fk∗∗0 =
e−i2πkj/(N/2) f2j .
j=0
119
III Orthogonality
Here k runs from 0 to 3 before the values start repeating. Similarly, F ∗00 is a transform of
length N/4 = 2 given by
N/4−1
X
Fk∗00 =
e−i2πkj/(N/4) f4j .
j=0
In this case k = 0, 1 and then the values repeat. Finally, the only j matching the pattern
010 is j = 2, so Fk010 is a transform of length one term given by
N/8−1
Fk010
=
X
−i2πkj/(N/8)
e
f2 =
0
X
e0 f2 . = f2
j=0
j=0
With this notation, the basic even–odd formula can be written
Fk∗∗∗ = Fk∗∗0 + ω kN Fk∗∗1 .
Recall that ωN = ei2π/N , so ω N = e−i2π/N .
Lets look at this equation when k = 0. We will represent the formula by the following
diagram.
0
F **
0
1
F **
0
120
ω0
N
F ***
0
III.5 The Discrete Fourier Transform
This diagram means that F0∗∗∗ is obtained by adding F0∗∗0 to ω 0N F0∗∗1 . (Of course ω0N = 1
so we could omit it.) Now lets add the diagrams for k = 1, 2, 3.
0
F **
0
0
F **
1
0
F **
2
0
F **
3
ω0
N
ω1
N
ω2
N
ω3
N
F ***
0
F ***
1
F ***
2
F ***
3
1
F **
0
1
F **
1
1
F **
2
1
F **
3
121
III Orthogonality
Now when we get to k = 4, we recall that F ∗∗0 and F ∗∗1 are discrete transforms of length
N/2 = 4. Therefore, by periodicity F4∗∗0 = F0∗∗0 , F5∗∗0 = F1∗∗0 , and so on. So in the formula
F4∗∗∗ = F4∗∗0 + ω 4N F4∗∗1 we may replace F4∗∗0 and F4∗∗1 with F0∗∗0 and F0∗∗1 respectively.
Making such replacements, we complete the first part of the diagram as follows.
0
F **
0
0
F **
1
0
F **
2
0
F **
3
1
F **
0
1
F **
1
1
F **
2
1
F **
3
122
ω0
N
ω1
N
ω2
N
ω3
N
ω4
N
ω5
N
ω6
N
ω7
N
F ***
0
F ***
1
F ***
2
F ***
3
F ***
4
F ***
5
F ***
6
F ***
7
III.5 The Discrete Fourier Transform
To move to the next level we analyze the discrete Fourier transforms on the left of this
diagram in the same way. This time we use the basic formula for the transform of length
N/2, namely
Fk∗∗0 = Fk∗00 + ω kN/2 Fk∗10
and
Fk∗∗1 = Fk∗01 + ωkN/2 Fk∗11 .
The resulting diagram shows how to go from the length two transforms to the final transform
on the right.
F*0 00
0
ωN/2
0
F **
0
F*100
ω1
0
F **
1
F*0 10
2
ωN/2
F*1 10
3
ωN/2
F *0 01
N/2
0
F **
2
ω0
N
ωN1
F ***
0
F ***
1
2
ωN
F ***
2
0
F **
3
ω3
N
F ***
3
0
ωN/2
1
F **
0
ω4
N
F ***
4
F *1 01
ω1
N/2
1
F **
1
5
ωN
F *0 11
ω2N/2
F *111
ω3N/2
1
F **
2
1
F **
3
ωN6
ωN7
F ***
5
F ***
6
F ***
7
123
III Orthogonality
Now we go down one more level. Each transform of length two can be constructed from
transforms of length one, i.e., from the original data in some order. We complete the diagram
as follows. Here we have inserted the value N = 8.
f0 = F 000
0
ω0
2
f4 = F100
0
ω1
2
f2 = F 010
0
ω 02
f6 = F110
0
ω12
f1 = F 001
0
ω0
2
f5 = F101
0
ω1
2
f3 = F 011
0
ω02
f7 = F 111
0
ω1
2
F*0 00
F*100
F*0 10
F*1 10
F *0 01
F *1 01
F *0 11
F *111
ω0
4
ω1
4
ω42
ω43
ω40
ω1
4
ω2
4
ω3
4
0
F **
0
0
F **
1
0
F **
2
0
F **
3
1
F **
0
1
F **
1
1
F **
2
1
F **
3
ω0
8
ω1
8
ω82
ω38
ω48
ω5
8
ω6
8
ω7
8
= c0
F ***
0
= c1
F ***
1
= c2
F ***
2
= c3
F ***
3
= c4
F ***
4
= c5
F ***
5
= c6
F ***
6
= c7
F ***
7
Notice that the fj ’s on the left of the diagram are in bit reversed order. In other words,
if we reverse the order of the bits in the binary expansion of the j’s, the resulting numbers
are ordered from 0 (000) to 7 (111).
Now we can describe the algorithm for the fast Fourier transform. Starting with the original
data [f0 , . . . , f7 ] we arrange the values in bit reversed order. Then we combine them pairwise,
as indicated by the left side of the diagram, to form the transforms of length 2. To do this we
we need to compute ω 2 = e−iπ = −1. Next we combine the transforms of length 2 according
to the middle part of the diagram to form the transforms of length 4. Here we use that
ω4 = e−iπ/2 = −i. Finally we combine the transforms of length 4 to obtain the transform of
length 8. Here we need ω 8 = e−iπ/4 = 2−1/2 − i2−1/2 .
The algorithm for values of N other than 8 is entirely analogous. For N = 2 or 4 we stop
at the first or second stage. For larger values of N = 2n we simply add more stages. How
many multiplications do we need to do? Well there are N = 2n multiplications per stage of
the algorithm (one for each circle on the diagram), and there are n = log2 (N ) stages. So the
number of multiplications is 2n n = N log2 (N )
As an example let us compute the discrete Fourier transform with N = 4 of the data
[f0 , f1 , f2 , f3 ] = [1, 2, 3, 4]. First we compute the bit reversed order of 0 = (00), 1 = (01), 2 =
(10), 3 = (11) to be (00) = 0, (10) = 2, (01) = 1, (11) = 3. We then do the rest of the
computation right on the diagram as follows.
124
III.5 The Discrete Fourier Transform
f0 =
1
1
1+3=4
1
4+6=10 = c 0
f2 =
3
−1
1−3=−2
−i
−2+2i
f1 =
2
1
2+4=6
−1
4−6=−2 = c 2
f3 =
4
−1
2−4=−2
i
−2−2i
= c1
= c3
The MATLAB/Octave command for computing the fast Fourier transform is fft. Let’s
verify the computation above.
> fft([1 2 3 4])
ans =
10 +
0i
-2 +
2i
-2 +
0i
-2 -
2i
The inverse fft is computed using ifft.
III.5.3 A frequency-amplitude plot for a sampled audio signal
Recall that a frequency-amplitude plot for the function y(t) defined on the interval [0, T ] is
a plot of the points (ωn , |cn |), where ωn = n/T and cn are the numbers appearing in the
Fourier series
∞
∞
X
X
cn e2πint/T
cn e2πiωn t =
y(t) =
n=−∞
n=−∞
If y(t) represents the sound of a musical instrument, then the frequency-amplitude plot gives
a visual representation of the strengths of the various frequencies present in the sound.
Of course, for an actual instrument there is no formula for y(t) and the best we can
do is to measure this function at a discrete set of points. To do this we pick a sampling
frequency Fs samples/second. Then we measure the function y(t) at times t = tj = j/Fs ,
j = 1, . . . N , where N = Fs T (so that tN = T ) and put the results yj = y(tj ) in a vector
y = [y1 , y2 , . . . , yN ]T . How can we make an approximate frequency-amplitude plot with this
information?
The key is to realize that the coefficients in the discrete Fourier transform of y can be used to
approximate the Fourier series coefficients cn . To see this, do a Riemann sum approximation
125
III Orthogonality
of the integral in the formula for cn . Using the equally spaced points tj with ∆tj = 1/Fs ,
and recalling that N = T Fs we obtain
1
cn =
T
Z
T
0
N
−1
X
e−2πint/T y(t)dt
≈
1
T
=
N −1
1 X −2πinj/(T Fs )
e
yj
T Fs
e−2πintj /T y(tj )∆tj
j=0
j=0
1
= c̃n
N
where c̃n is the nth coefficient in the discrete Fourier transform of y
The frequency corresponding to cn is ωn = n/T = nFs /N . So, for an approximate
frequency-amplitude plot, we can plot the points (nFs /N, |c̃n |/N ).
However, it is important to realize that the approximation cn ≈ c̃n /N is only good for small
n. The reason is that the Riemann sum will do a worse job in approximating the integral
when the integrand is oscillating rapidly, that is, when n is large. So we should only plot a
restricted range of n. In fact, it never makes sense to plot more than N/2 points. The reason
for this is c̃n+N = c̃n and, for y real valued, c̃−n = c̃n . These facts imply that |c̃n | = |c̃N −n |,
so that the values of |c̃n | in the range [0, N/2 − 1] are the same as the values in [N/2, N − 1],
with the order reversed.
To compare the meanings of the coefficients cn and c̃n it is instructive to consider the
formulas (both exact) for the Fourier series and the discrete Fourier transform for yj = y(tj ):
yj =
y(tj ) =
N −1
1 X
c̃n e2πinj/N
N n=0
∞
X
n=−∞
cn e2πintj /T =
∞
X
cn e2πinj/N
n=−∞
The coefficients cn and c̃n /N are close for n close to 0, but then their values diverge so that
the infinite sum and the finite sum above both give the same answer.
Now let’s try and make a frequency amplitude plot using MATLAB/Octave for a sampled
flute contained in the audio file F6.baroque.au available at
http://www.phys.unsw.edu.au/music/flute/baroque/sounds/F6.baroque.au.
This file contains a sampled baroque flute playing the note F6 , which has a frequency of
1396.91 Hz. The sampling rate is Fs = 22050 samples/second.
Audio processing is one area where MATLAB and Octave are different. The Octave code
to load the file F6.baroque.au is
126
III.5 The Discrete Fourier Transform
y=loadaudio(’F6.baroque’,’au’,8);
while the MATLAB code is
y=auread(’F6.baroque.au’);
After this step the sampled values are loaded in the vector y. Now we compute the FFT
of y and store the resulting values c̃n in a vector tildec. Then we compute a vector omega
containing the frequencies and make a plot of these frequencies against |c̃n |/N . We plot the
first Nmax=N/4 values.
tildec = fft(y);
N=length(y);
Fs=22050;
omega=[0:N-1]*Fs/N;
Nmax=floor(N/4);
plot(omega(1:Nmax), abs(tildec(1:Nmax)/N));
Here is the result.
12
10
8
6
4
2
0
0
1000
2000
3000
4000
5000
Notice the large spike at ω ≈ 1396 corresponding to the note F6 . Smaller spikes appear at
the overtone series, but evidently these are quite small for a flute.
127
Chapter IV
Eigenvalues and Eigenvectors
129
IV Eigenvalues and Eigenvectors
IV.1 Eigenvalues and Eigenvectors
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the definition of eigenvalues and eigenvectors and be able to compute them
using the standard procedure.
• use MATLAB/Octave commands poly and root to compute the characteristic polynomial and its roots, and eig to compute the eigenvalues and eigenvectors.
• write down the definitions of algebraic and geometric multiplicities of eigenvectors when
there are repeated eigenvalues.
• use eigenvalues and eigenvectors to perform matrix diagonalization.
• recognize the form of the Jordan Canonical Form for non-diagonalizable matrices.
• explain the relationship between eigenvalues and the determinant and trace of a matrix.
• use eigenvalues to compute powers of a diagonalizable matrix.
IV.1.1 Definition
Let A be an n × n matrix. A number λ and non-zero vector v are an eigenvalue eigenvector
pair for A if
Av = λv
Although v is required to be nonzero, λ = 0 is possible. If v is an eigenvector, so is sv for
any number s 6= 0.
Rewrite the eigenvalue equation as
(λI − A)v = 0
Then we see that v is a non-zero vector in the nullspace N (λI − A). Such a vector only
exists if λI − A is a singular matrix, or equivalently if
det(λI − A) = 0
130
IV.1 Eigenvalues and Eigenvectors
IV.1.2 Standard procedure
This leads to the standard textbook method of finding eigenvalues. The function of λ defined
by p(λ) = det(λI −A) is a polynomial of degree n, called the characteristic polynomial, whose
zeros are the eigenvalues. So the standard procedure is:
• Compute the characteristic polynomial p(λ)
• Find all the zeros (roots) of p(λ). This is equivalent to completely factoring p(λ) as
p(λ) = (λ − λ1 )(λ − λ2 ) · · · (λ − λn )
Such a factorization always exists if we allow the possibility that the zeros λ1 , λ2 , . . .
are complex numbers. But it may be hard to find. In this factorization there may
be repetitions in the λi ’s. The number of times a λi is repeated is called its algebraic
multiplicity.
• For each distinct λi find N (λI − A), that is, all the solutions to
(λi I − A)v = 0
The non-zero solutions are the eigenvectors for λi .
IV.1.3 Example 1
This is the typical case where all the eigenvalues are distinct. Let
3 −6 −7
A= 1
8
5
−1 −2 1
Then, expanding the determinant, we find
det(λI − A) = λ3 − 12λ2 + 44λ − 48
This can be factored as
λ3 − 12λ2 + 44λ − 48 = (λ − 2)(λ − 4)(λ − 6)
So the eigenvalues are 2, 4 and 6.
These steps can be done with MATLAB/Octave using poly and root. If A is a square
matrix, the command poly(A) computes the characteristic polynomial, or rather, its coefficients.
131
IV Eigenvalues and Eigenvectors
> A=[3 -6 -7; 1 8 5; -1 -2 1];
> p=poly(A)
p =
1.0000
-12.0000
44.0000
-48.0000
Recall that the coefficient of the highest power comes first. The function roots takes as
input a vector representing the coefficients of a polynomial and returns the roots.
>roots(p)
ans =
6.0000
4.0000
2.0000
To find the eigenvector(s) for λ1 = 2 we must solve the homogeneous equation (2I −A)v = 0.
Recall that eye(n) is the n × n identity matrix I
>rref(2*eye(3) - A)
ans =
1
0
0
0
1
0
-1
1
0
From this we can read off the solution
1
v1 = −1
1
Similarly we find for λ2 = 4 and λ3 = 6 that the corresponding eigenvectors are
−1
−2
v2 = −1
v3 = 1
1
0
The three eigenvectors v1 , v2 and v3 are linearly independent and form a basis for R3 .
The MATLAB/Octave command for finding eigenvalues and eigenvectors is eig. The
command eig(A) lists the eigenvalues
132
IV.1 Eigenvalues and Eigenvectors
>eig(A)
ans =
4.0000
2.0000
6.0000
while the variant [V,D] = eig(A) returns a matrix V whose columns are eigenvectors and a
diagonal matrix D whose diagonal entries are the eigenvalues.
>[V,D] = eig(A)
V =
5.7735e-01
5.7735e-01
-5.7735e-01
5.7735e-01
-5.7735e-01
5.7735e-01
-8.9443e-01
4.4721e-01
2.2043e-16
D =
4.00000
0.00000
0.00000
0.00000
2.00000
0.00000
0.00000
0.00000
6.00000
Notice that the eigenvectors have been normalized to have length one. Also, since they
have been computed numerically, they are not exactly correct. The entry 2.2043e-16 (i.e.,
2.2043 × 10−16 ) should actually be zero.
IV.1.4 Example 2
This example has a repeated eigenvalue.
The characteristic polynomial is
1 1 0
A = 0 2 0
0 −1 1
det(λI − A) = λ3 − 4λ2 + 5λ − 2 = (λ − 1)2 (λ − 2)
In this example the eigenvalues are 1 and 2, but the eigenvalue 1 has algebraic multiplicity
2.
133
IV Eigenvalues and Eigenvectors
Let’s find the eigenvector(s) for λ1 = 1 we have
0 1 0
I − A = 0 1 0
0 −1 0
From this it is easy to see that there are two linearly independent eigenvectors for this
eigenvalue:
0
1
v1 = 0 and w1 = 0
1
0
In this case we say that the geometric multiplicity is 2. In general, the geometric multiplicity
is the number of independent eigenvectors, or equivalently the dimension of N (λI − A)
The eigenvalue λ2 = 2 has eigenvector
−1
v2 = −1
1
So, although this example has repeated eigenvalues, there still is a basis of eigenvectors.
IV.1.5 Example 3
Here is an example where the geometric multiplicity is less than the algebraic multiplicity. If
2 1 0
A = 0 2 1
0 0 2
then the characteristic polynomial is
det(λI − A) = (λ − 2)3
so there is one eigenvalue λ1 = 2 with algebraic multiplicity 3.
To find the eigenvectors we compute
0 −1 0
2I − A = 0 0 −1
0 0
0
From this we see that there is only one independent solution
1
v1 = 0
0
134
IV.1 Eigenvalues and Eigenvectors
Thus the geometric multiplicity dim(N (2I − A)) is 1. What does MATLAB/Octave do in
this situation?
>A=[2 1 0; 0 2 1; 0 0 2];
>[V D] = eig(A)
V =
1.00000
0.00000
0.00000
-1.00000
0.00000
0.00000
1.00000
-0.00000
0.00000
D =
2
0
0
0
2
0
0
0
2
It simply returned the same eigenvector three times.
In this example, there does not exist a basis of eigenvectors.
IV.1.6 Example 4
Finally, here is an example where the eigenvalues are complex, even though the matrix has
real entries. Let
0 −1
A=
1 0
Then
det(λI − A) = λ2 + 1
which has no real roots. However
λ2 + 1 = (λ + i)(λ − i)
so the eigenvalues are λ1 = i and λ2 = −i. The eigenvectors are found with the same
procedure as before, except that now we must use complex arithmetic. So for λ1 = i we
compute
i 1
iI − A =
−1 i
There is trick for computing the null space of a singular 2 × 2 matrix. Since the two rows
must be multiples of each other (in this case the second row is i times the first row) we simply
135
IV Eigenvalues and Eigenvectors
a
need to find a vector
with ia + b = 0. This is easily achieved by flipping the entries in
b
the first row and changing the sign of one of them. Thus
1
v1 =
−i
If a matrix has real entries, then the eigenvalues and eigenvectors occur in conjugate pairs.
This can be seen directly from the eigenvalue equation Av = λv. Taking complex conjugates
(and using that the conjugate of a product is the product of the conjugates) we obtain
Āv̄ = λ̄v̄ But if A is real then Ā = A so Av̄ = λ̄v̄, which shows that λ̄ and v̄ are also an
eigenvalue eigenvector pair.
From this discussion it follows that v2 is the complex conjugate of v1
1
v2 =
i
IV.1.7 A basis of eigenvectors
In three of the four examples above the matrix A had a basis of eigenvectors. If all the
eigenvalues are distinct, as in the first example, then the corresponding eigenvectors are
always independent and therefore form a basis.
To see why this is true, suppose A has eigenvalues λ1 , . . . , λn that are all distinct, that is,
λi =
6 λj for i 6= j. Let v1 , . . . , vn be the corresponding eigenvectors.
Now, starting with the first two eigenvectors, suppose a linear combination of them equals
zero:
c1 v1 + c2 v2 = 0
Multiplying by A and using the fact that these are eigenvectors, we obtain
c1 Av1 + c2 Av2 = c1 λ1 v1 + c2 λ2 v2 = 0
On the other hand, multiplying the original equation by λ2 we obtain
c1 λ2 v1 + c2 λ2 v2 = 0.
Subtracting the equations gives
c1 (λ2 − λ1 )v1 = 0
Since (λ2 − λ1 ) 6= 0 and, being an eigenvector, v1 6= 0 it must be that c1 = 0. Now returning
to the original equation we find c2 v2 = 0 which implies that c2 = 0 too. Thus v1 and v2 are
linearly independent.
Now let’s consider three eigenvectors v1 , v2 and v3 . Suppose
c1 v1 + c2 v2 + c3 v3 = 0
136
IV.1 Eigenvalues and Eigenvectors
As before, we multiply by A to get one equation, then multiply by λ3 to get another equation.
Subtracting the resulting equations gives
c1 (λ1 − λ3 )v1 + c2 (λ2 − λ3 )v2 = 0
But we already know that v1 and v2 are independent. Therefore c1 (λ1 −λ3 ) = c2 (λ2 −λ3 ) = 0.
Since λ1 − λ3 6= 0 and λ2 − λ3 6= 0 this implies c1 = c2 = 0 too. Therefore v1 , v2 and v3 are
independent.
Repeating this argument, we eventually find that all the eigenvectors v1 , . . . , vn are independent.
In example 2 above, we saw that it might be possible to have a basis of eigenvectors even
when there are repeated eigenvalues. For some classes of matrices (for example symmetric
matrices (AT = A) or orthogonal matrices) a basis of eigenvectors always exists, whether or
not there are repeated eigenvalues. Will will consider this in more detail later in the course.
IV.1.8 When there are not enough eigenvectors
Let’s try to understand a little better the exceptional situation where there are not enough
eigenvectors to form a basis. Consider
1
1
Aǫ =
0 1+ǫ
1
when ǫ = 0 this matrix has a single eigenvalues λ = 1 and only one eigenvector v1 =
.
0
What happens when we change ǫ slightly? Then the eigenvalues change to 1 and 1 + ǫ, and
being distinct, they must have independent eigenvectors. A short calculation reveals that
they are
1
1
v1 =
v2 =
0
ǫ
These two eigenvectors are almost, but not quite, dependent. When ǫ becomes zero they
collapse and point in the same direction.
In general, if you start with a matrix with repeated eigenvalues and too few eigenvectors,
and change the entries of the matrix a little, some of the eigenvectors (the ones corresponding
to the eigenvalues whose algebraic multiplicity is higher than the geometric multiplicity) will
split into several eigenvectors that are almost parallel.
IV.1.9 Diagonalization
Suppose A is an n × n matrix with eigenvalues λ1 , · · · , λn and a basis of eigenvectors
v1 , . . . , vn . Form the matrix with eigenvectors as columns
S = v1 v2 · · · vn
137
IV Eigenvalues and Eigenvectors
Then
AS = Av1 Av2 · · · Avn
= λ1 v1 λ2 v2 · · · λn vn
λ1 0 0
0 λ2 0
= v1 v2 · · · vn 0 0 λ3
..
.
0 0 0
= SD
···
···
···
···
0
0
0
..
.
λn
where D is the diagonal matrix with the eigenvalues on the diagonal. Since the columns of
S are independent, the inverse exists and we can write
A = SDS −1
S −1 AS = D
This is called diagonalization.
Notice that the matrix S is exactly the one returns by the MATLAB/Octave call [S D] = eig(A).
>A=[1 2 3; 4 5 6; 7 8 9];
>[S D] = eig(A);
>S*D*S^(-1)
ans =
1.0000
4.0000
7.0000
2.0000
5.0000
8.0000
3.0000
6.0000
9.0000
IV.1.10 Jordan canonical form
If A is a matrix that cannot be diagonalized, there still exits a similar factorization called
the Jordan Canonical Form. It turns out that any matrix A can be written as
A = SBS −1
where B is a block diagonal matrix. The
B1
0
B=0
..
.
0
138
matrix B has the form
0
0 ··· 0
B2 0 · · · 0
0 B3 · · · 0
..
.
0
0
···
Bk
IV.1 Eigenvalues and Eigenvectors
Where each submatrix Bi (called a Jordan block) has
and 1’s on the superdiagonal.
λi 1 0 · · ·
0 λi 1 · · ·
Bi = 0 0 λi · · ·
..
.
0
0
0
···
a single eigenvalue on the diagonal
0
0
0
..
.
λi
If all the blocks are of size 1 × 1 then there are no 1’s and the matrix is diagonalizable.
IV.1.11 Eigenvalues, determinant and trace
Recall that the determinant satisfies det(AB) = det(A) det(B) and det(S −1 ) = 1/ det(S).
Also, the determinant of a diagonal matrix (or more generally of an upper triangular matrix)
is the product of the diagonal entries. Thus if A is diagonalizable then
det(A) = det(SDS −1 ) = det(S) det(D) det(S −1 ) = det(S) det(D)/ det(S) = det(D) = λ1 λ2 · · · λn
Thus the determinant of a matrix is the product of the eigenvalues. This is true for nondiagonalizable matrices as well, as can be seen from the Jordan Canonical Form. Notice that
the number of times a particular λi appears in this product is the algebraic multiplicity of
that eigenvalues.
P
The trace of a matrix is the sum of the diagonal entries. If A = [ai,j ] then tr(A) = i ai,i .
Even though it is not true that AB = BA in general, the trace is not sensitive to the change
in order:
X
X
tr(AB) =
ai,j bj,i =
bj,i ai,j = tr(BA)
i,j
i,j
Thus (taking A = SD and B = S −1 )
tr(A) = tr(SDS −1 ) = tr(S −1 SD) = tr(D) = λ1 + λ2 + · · · + λn
Thus the trace of a diagonalizable matrix is the sum of the eigenvalues. Again, this is true
for non-diagonalizable matrices as well, and can be seen from the Jordan Canonical Form.
IV.1.12 Powers of a diagonalizable matrix
If A is diagonalizable then its powers Ak are easy to compute.
Ak = SDS −1 SDS −1 SDS −1 · · · SDS −1 = SD k S −1
139
IV Eigenvalues and Eigenvectors
because all of the factors S −1 S cancel. Since powers of the diagonal matrix D are given by
k
λ1 0 0 · · · 0
0 λk 0 · · · 0
2
k
Dk = 0 0 λ3 · · · 0
..
..
.
.
0 0 0 · · · λkn
this formula provides an effective way to understand and compute Ak for large k.
140
IV.2 Power Method for Computing Eigenvalues
IV.2 Power Method for Computing Eigenvalues
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the properties of the eigenvalues and eigenvectors of real symmetric matrices.
• write down the definition and properties of Hermitian matrices.
• use the power method to compute the eigenvalue/eigenvector of a Hermitian matrix,
where the eigenvalue is closest to a given number.
IV.2.1 Eigenvalues of real symmetric matrices
If A is real (that is, the entries are all real numbers) and symmetric (that is AT = A) then
the eigenvalues of A are all real, and the eigenvectors can be chosen to form an orthonormal
basis.
To see that the eigenvalues must be real, let’s start with an eigenvalue eigenvector pair λ,
v for A. For the moment, we allow the possibility that λ and v are complex.
Since A is real and symmetric we have
hv, Avi = hAv, vi
and since Av = λv this implies
hv, λvi = hλv, vi
T
Here we are using the inner product for complex vectors given by h0, wi = 0 w. This means
that the λ on the right side is conjugated, that is,
λhv, vi = λhv, vi.
Since v is an eigenvector, it cannot be zero. So hv, vi = kvk2 6= 0. Therefore we may divide
by hv, vi to conclude
λ = λ.
This shows that λ is real.
Now let’s show that eigenvectors corresponding to two distinct eigenvalues must be orthogonal. If Av1 = λ1 v1 and Av2 = λ2 v2 with λ1 6= λ2 , then starting with the equation that
follows from the symmetry of A
hAv1 , v2 i = hv1 , Av2 i
141
IV Eigenvalues and Eigenvectors
we find
λ1 hv1 , v2 i = λ2 hv1 , v2 i
Here λ1 should appear as λ1 , but we already know that eigenvalues are real so λ1 = λ1 . This
can be written
(λ1 − λ2 )hv1 , v2 i = 0
and since λ1 − λ2 6= 0 this implies
hv1 , v2 i = 0
This calculation shows that if A has distinct eigenvalues then the eigenvectors are all
orthogonal, and by rescaling them, we can obtain an orthonormal basis of eigenvectors.
In fact, even it A has repeated eigenvalues, it is still true that an orthonormal basis of
eigenvectors exists.
If A is real and symmetric, then the eigenvectors can be chosen to be real. One way to
see this is to notice that if once we know that λ is real then all the calculations involved in
computing the nullspace of λI − A only involve real numbers. This implies that the matrix
that diagonalizes A can be chosen to be an orthogonal matrix.
T
If A has complex entries, but satisfies A∗ = A it is called Hermitian. (Recall that A∗ = A .)
The argument above still is valid for Hermitian matrices and shows that all the eigenvalues
are real. There also exists an orthonormal basis of eigenvectors. However, in contrast to
the case where A is real and symmetric, the eigenvectors may have complex entries. Thus a
Hermitian matrix may be diagonalized by a unitary matrix.
If A is any matrix with real entries, then A + AT is real symmetric. (The matrix AT A is
also real symmetric, and has the additional property that all the eigenvalues are positive.)
We can use this to produce random symmetric matrices in MATLAB/Octave like this:
>A = rand(4,4);
>A = A+A’
A =
0.043236
1.240654
0.658890
0.437168
1.240654
1.060615
0.608234
0.911889
0.658890
0.608234
1.081767
0.706712
0.437168
0.911889
0.706712
1.045293
Let’s check the eigenvalues and vectors or A
>[V, D] = eig(A)
V =
142
IV.2 Power Method for Computing Eigenvalues
-0.81345
0.54456
0.15526
-0.13285
0.33753
0.19491
0.35913
-0.84800
0.23973
0.55585
-0.78824
-0.11064
0.40854
0.59707
0.47497
0.50100
0.00000
0.36240
0.00000
0.00000
0.00000
0.00000
0.55166
0.00000
0.00000
0.00000
0.00000
3.15854
D =
-0.84168
0.00000
0.00000
0.00000
The eigenvalues are real, as expected. Also, the eigenvectors contained in the columns of the
matrix V have been normalized. Thus V is orthogonal:
>V’*V
ans =
1.0000e+00
9.1791e-17
-1.4700e-16
1.3204e-16
6.5066e-17
1.0000e+00
-1.0012e-16
2.2036e-17
-1.4700e-16
-1.0432e-16
1.0000e+00
-1.0330e-16
1.4366e-16
2.2036e-17
-1.2617e-16
1.0000e+00
(at least, up to numerical error.)
IV.2.2 The power method
The power method is a very simple method for finding a single eigenvalue–eigenvector pair.
Suppose A is an n × n matrix. We assume that A is real symmetric, so that all the
eigenvalues are real. Now let x0 be any vector of length n. Perform the following steps:
• Multiply by A
• Normalize to unit length.
repeatedly. This generates a series of vectors x0 , x1 , x2 , . . .. It turns out that these vectors
converge to the eigenvector corresponding to the eigenvalue whose absolute value is the
largest.
143
IV Eigenvalues and Eigenvectors
To verify this claim, let’s first find a formula for xk . At each stage of this process, we are
multiplying by A and then by some number. Thus xk must be a multiple of Ak x0 . Since the
resulting vector has unit length, that number must be 1/kAk x0 k. Thus
xk =
Ak x0
kAk x0 k
We know that A has a basis of eigenvectors v1 , v2 , . . . , vn . Order them so that |λ1 | > |λ2 | ≥
· · · ≥ |λn |. (We are assuming here that |λ1 | =
6 |λ2 |, otherwise the power method runs into
difficulty.) We may expand our initial vector x0 in this basis
x0 = c1 v1 + c2 v2 + · · · cn vn
We need that c1 6= 0 for this method to work, but if x0 is chosen at random, this is almost
always true.
Since Ak vi = λki vi we have
Ak x0 = c1 λk1 v1 + c2 λk2 v2 + · · · cn λkn vn
= λk1 c1 v1 + c2 (λ2 /λ1 )k v2 + · · · cn (λn /λ1 )k vn
= λk1 (c1 v1 + ǫk )
where ǫk → 0 as k → ∞. This is because |(λi /λ1 )| < 1 for every i > 1 so the powers tend to
zero. Thus
kAk x0 k = |λ1 |k kc1 v1 + ǫk k
so that
Ak x0
kAk x0 k
λ1 k c1 v1 + ǫk
=
|λ1 |
kc1 v1 + ǫk k
v1
k
→ (±) ±
kv1 k
xk =
We have shown that xk converges, except for a possible sign flip at each stage, to a normalized eigenvector corresponding to λ1 . The sign flip is present exactly when λ1 < 0. Knowing
v1 (or a multiple of it) we can find λ1 with
λ1 =
hv1 , Av1 i
kv1 k2
This gives a method for finding the largest eigenvalue (in absolute value) and the corresponding eigenvector. Let’s try it out.
144
IV.2 Power Method for Computing Eigenvalues
>A = [4 1 3;1 3 2; 3 2 5];
>x=rand(3,1);
>for k = [1:10]
>y=A*x;
>x=y/norm(y)
>end
x =
0.63023
0.38681
0.67319
x =
0.58690
0.37366
0.71828
x =
0.57923
0.37353
0.72455
x =
0.57776
0.37403
0.72546
x =
0.57745
0.37425
0.72559
x =
0.57738
0.37433
0.72561
x =
145
IV Eigenvalues and Eigenvectors
0.57736
0.37435
0.72562
x =
0.57735
0.37436
0.72562
x =
0.57735
0.37436
0.72562
x =
0.57735
0.37436
0.72562
This gives the eigenvector. We compute the eigenvalue with
>x’*A*x/norm(x)^2
ans =
8.4188
Let’s check:
>[V D] = eig(A)
V =
0.577350
0.441225
-0.687013
0.577350
-0.815583
-0.038605
0.577350
0.374359
0.725619
D =
1.19440
0.00000
0.00000
146
0.00000
2.38677
0.00000
0.00000
0.00000
8.41883
IV.2 Power Method for Computing Eigenvalues
As expected, we have computed the largest eigenvalue and eigenvector. Of course, a serious
program that uses this method would not just iterate a fixed number (above it was 10) times,
but check for convergence, perhaps by checking whether kxk −xk−1 k was less than some small
number, and stopping when this was achieved.
So far, the power method only computes the eigenvalue with the largest absolute value, and
the corresponding eigenvector. What good is that? Well, it turns out that with an additional
twist we can compute the eigenvalue closest to any number s. The key observation is that
the eigenvalues of (A − sI)−1 are exactly (λi − s)−1 (unless, of course, A − sI is not invertible.
But then s is an eigenvalue itself and we can stop looking.) Moreover, the eigenvectors of A
and (A − sI)−1 are the same.
Let’s see why this is true. If
Av = λv
then
(A − sI)v = (λ − s)v.
Now if we multiply both sides by (A − sI)−1 and divide by λ − s we get
(λ − s)−1 v = (A − sI)−1 v.
These steps can be run backwards to show that if (λ − s)−1 is an eigenvalue of (A − sI)−1
with eigenvector v, then λ is an eigenvalue of A with the same eigenvector.
Now start with an arbitrary vector x0 and define
xk+1 =
(A − sI)−1 xk
.
k(A − sI)−1 xk k
Then xk will converge to the eigenvector vi of (A − sI)−1 for which |λi − s|−1 is the largest.
But, since the eigenvectors of A and A − sI are the same, vi is also an eigenvector of A. And
since |λi − s|−1 is largest when λi is closest to s, we have computed the eigenvector vi of A
for which λi is closest to s. We can now compute λi by comparing Avi with vi
Here is a crucial point: when computing (A − sI)−1 xk in this procedure, we should not
actually compute the inverse. We don’t need to know the whole matrix (A − sI)−1 , but just
the vector (A − sI)−1 xk . This vector is the solution y of the linear equation (A − sI)y = xk .
In MATLAB/Octave we would therefore use something like (A - s*eye(n))\Xk.
Let’s try to compute the eigenvalue of the matrix A above closest to 3.
>A = [4 1 3;1 3 2; 3 2 5];
>x=rand(3,1);
>for k = [1:10]
>y=(A-3*eye(3))\x;
>x=y/norm(y)
>end
147
IV Eigenvalues and Eigenvectors
x =
0.649008
-0.756516
0.080449
x =
-0.564508
0.824051
0.047657
x =
0.577502
-0.815593
-0.036045
x =
-0.576895
0.815917
0.038374
x =
0.577253
-0.815659
-0.038454
x =
-0.577311
0.815613
0.038562
x =
0.577338
-0.815593
-0.038590
x =
148
IV.2 Power Method for Computing Eigenvalues
-0.577346
0.815587
0.038600
x =
0.577349
-0.815585
-0.038603
x =
-0.577350
0.815584
0.038604
This gives the eigenvector. Now we can find the eigenvalue
> lambda = x’*A*x/norm(x)^2
lambda = 2.3868
Comparing with the results of eig above, we see that we have computed the middle eigenvalue
and eigenvector.
149
IV Eigenvalues and Eigenvectors
IV.3 Recursion Relations
Prerequisites and Learning Goals
After completing this section, you should be able to
• use matrix equations to solve a recurrence relation, for example the relation defining
Fibonacci numbers.
• determine initial values for which the solution of a recurrence relation will become large
or small (depending of the eigenvalues of the associated matrix).
IV.3.1 Fibonacci numbers
Consider the sequence of numbers given by a multiple of powers of the golden ratio
√ !n
1
1+ 5
√
n = 1, 2, 3 . . . .
2
5
When n is large, these numbers are almost integers:
>format long;
>((1+sqrt(5))/2)^30/sqrt(5)
ans =
832040.000000241
>((1+sqrt(5))/2)^31/sqrt(5)
ans =
1346268.99999985
>((1+sqrt(5))/2)^32/sqrt(5)
ans =
2178309.00000009
Why? To answer this question we introduce the Fibonacci sequence:
0 1 1
2 3 4
5 8 13
...
where each number in the sequence is obtained by adding the previous two. If you go far
enough along in this sequence you will encounter
...
150
832040
1346269
2178309
...
IV.3 Recursion Relations
and you can check (without using MATLAB/Octave, I hope) that the third number is the
sum of the previous two.
But why should powers of the golden ratio be very nearly, but not quite, equal to Fibonacci
numbers? The reason is that the Fibonacci sequence is defined by a recursion relation. For
the Fibonacci sequence F0 , F1 , F2 , . . . the recursion relation is
Fn+1 = Fn + Fn−1
This equation, together with the identity Fn = Fn can be written in matrix form as
1 1
Fn+1
Fn
=
Fn
1 0 Fn−1
Thus, taking n = 1, we find
1 1 F1
F2
=
1 0 F0
F1
Similarly
2 F1
1 1
F3
1 1 F2
=
=
1 0
F0
1 0 F1
F2
and continuing like this we find
n 1 1
Fn+1
F1
=
Fn
1 0
F0
Finally, since F0 = 0 and F1 = 1 we can write
n 1
Fn+1
1 1
=
0
1 0
Fn
We can diagonalize the matrix
to get a formula for the Fibonacci numbers. The eigenvalues
1 1
are
and eigenvectors of
1 0
√
1+ 5
λ1 =
2
and
√
1− 5
λ2 =
2
v1 =
"
√ #
1+ 5
2
v2 =
"
√ #
1− 5
2
1
1
This implies
λ1 λ2
1 1
−1
n
1 λ1 λ2 λn1 0
λ1 λ2
λ1 0
1 −λ2
=√
0 λn2
1 1
0 λn2 −1 λ1
5 1 1
151
IV Eigenvalues and Eigenvectors
so that
1 λn+1
1 λ1 λ2 λn1 0
1 −λ2 1
Fn+1
− λn+1
1
2
=√
=√
0
0 λn2 −1 λ1
Fn
λn1 − λn2
5 1 1
5
In particular
1
Fn = √ (λn1 − λn2 )
5
Since λ2 ∼ −0.6180339880 is smaller than 1 in absolute value, the powers λn2 become small
very quickly as n becomes large. This explains why
1
Fn ∼ √ λn1
5
for large n.
If we want to use MATLAB/Octave to compute Fibonacci numbers, we don’t need to
bother diagonalizing the matrix.
>[1 1;1 0]^30*[1;0]
ans =
1346269
832040
produces the same Fibonacci numbers as above.
IV.3.2 Other recursion relations
The idea that was used to solve for the Fibonacci numbers can be used to solve other recursion
relations. For example the three-step recursion
xn+1 = axn + bxn−1 + cxn−2
can be written as a matrix equation
xn+1
a b c
xn
xn = 1 0 0 xn−1
xn−1
0 1 0 xn−2
so given three initial values x0 , x1 and x2 we can find the rest by computing powers of a
3 × 3 matrix.
In the next section we will solve a recurrence relation that arises in Quantum Mechanics.
152
IV.4 The Anderson Tight Binding Model
IV.4 The Anderson Tight Binding Model
Prerequisites and Learning Goals
After completing this section, you should be able to
• describe a bound state with energy E for the discrete Schrodinger equation for a single
electron moving in a one dimensional semi-infinite crystal.
• describe a scattering state with energy E.
• compute the energies for which a bound state exists and identify the conduction band,
for a potential that has only one non-zero value.
• compute the conduction bands for a one dimensional crystal.
IV.4.1 Description of the model
Previously we studied how to approximate differential equations by matrix equations. If we
apply this discretization procedure to the Schrödinger equation for an electron moving in a
solid we obtain the Anderson tight binding model.
We will consider a single electron moving in a one dimensional semi-infinite crystal. The
electron is constrained to live at discrete lattice points, numbered 0, 1, 2, 3, . . .. These can be
thought of as the positions of the atoms. For each lattice point n there is a potential Vn that
describes how much the atom at that lattice point attracts or repels the electron. Positive
Vn ’s indicate repulsion, whereas negative Vn ’s indicate attraction. Typical situations studied
in physics are where the Vn ’s repeat the same pattern periodically (a crystal), or where they
are chosen at random (disordered media). In fact, the term Anderson model usually refers
to the random case, where the potentials are chosen at random, independently for each site.
The wave function for the electron is a sequence of complex numbers Ψ = {ψ0 , ψ1 , ψ2 , . . .}.
The sequence Ψ is called a bound state with energy E if satisfies the following three conditions:
(1) The discrete version of the time independent Schrödinger equation
−ψn+1 − ψn−1 + Vn ψn = Eψn ,
(2) the boundary condition
ψ0 = 0,
(3) and the normalization condition
2
N =
∞
X
n=0
|ψn |2 < ∞.
153
IV Eigenvalues and Eigenvectors
This conditions are trivially satisfies if Ψ = {0, 0, 0, . . .} so we specifically exclude this case.
(In fact Ψ is actually the eigenvector of an infinite matrix so this is just the condition that
eigenvectors must be non-zero.)
Given an energy E, it is always possible to find a wave function Ψ to satisfy conditions
(1) and (2). However for most energies E, none of these Ψ’s will be getting small for large
n, so the normalization condition (3) will not be satisfied. There are only a discrete set of
energies E for which a bound state satisfying all three conditions is satisfied. In other words,
the energy E is quantized.
If E is one of the allowed energy values and Ψ is the corresponding bound state, then the
numbers |ψn |2 /N 2 are interpreted as the probabilities of finding an electron with energy E at
the nth site. These numbers add up to 1, consistent with the interpretation as probabilities.
Notice that if Ψ is a bound state with energy E, then so is any non-zero multiple aΨ =
{aψ0 , aψ1 , aψ2 , . . .}. Replacing Ψ with aΨ has no effect on the probabilities because N
changes to aN , so the a’s cancel in |ψn |2 /N 2 .
IV.4.2 Recursion relation
The discrete Schrödinger equation (1) together with the initial condition (2) is a recursion
relation that can be solved using the method we saw in the previous section. We have
ψn
Vn − E −1
ψn+1
=
1
0
ψn−1
ψn
so if we set
ψ
xn = n+1
ψn
and
z −1
A(z) =
1 0
then this implies
xn = A(Vn − E)A(Vn−1 − E) · · · A(V1 − E)x0 .
Condition (2) says that
x0 =
ψ1
,
0
since ψ0 = 0.
In fact, we may assume ψ1 = 1, since replacing Ψ with aΨ where a = 1/ψ1 results in a
bound state where this is true. Dividing by ψ1 is possible unless ψ1 = 0. But if ψ1 = 0 then
x0 = 0 and the recursion implies that every ψk = 0. This is not an acceptable bound state.
Thus we may assume
1
x0 =
0
154
IV.4 The Anderson Tight Binding Model
So far we are able to compute xn (and thus ψn ) satisfying conditions (1) and (2) for any
values of V1 , V2 , . . .. Condition (3) is a statement about the large n behaviour of ψn . This
can be very difficult to determine, unless we know more about the values Vn .
IV.4.3 A potential with most values zero
We will consider the very simplest situation where V1 = −a and all the other Vn ’s are equal
to zero. Let us try to determine for what energies E a bound state exists.
In this situation
xn = A(−E)n−1 A(−a − E)x0 = A(−E)n−1 x1
where
−a − E −1 1
−(a + E)
x1 = A(−a − E)x0 =
=
1
0
0
0
The large n behavior of xn can be computed using the eigenvalues and eigenvectors of A(−E).
Suppose they are λ1 , v1 and λ2 , v2 . Then we expand
x1 = a1 v1 + a2 v2
and conclude that
xn = A(−E)n−1 (a1 v1 + a2 v2 ) = a1 λ1n−1 v1 + a2 λ2n−1 v2
Keep in mind that all the quantities in this equation depend on E. Our goal is to choose E
so that the xn become small for large n.
Before computing the eigenvalues of A(−E), let’s note that det(A(−E)) = 1. This implies
that λ1 λ2 = 1
Suppose the eigenvalues are complex. Then, since A(−E) has real entries, they must be
complex conjugates. Thus λ2 = λ1 and 1 = λ1 λ2 = λ1 λ1 = |λ1 |2 . This means that λ1 and
λ2 lie on the unit circle in the complex plane. In other words, λ1 = eiθ and λ2 = e−iθ for
some θ. This implies that λ1n−1 = ei(n−1)θ is also on the unit circle, and is not getting small.
Similarly λ2n−1 is not getting small. So complex eigenvalues will never lead to bound states.
In fact, complex eigenvalues correspond to scattering states, and the energy values for which
eigenvalues are complex are the energies at which the electron can move freely through the
crystal.
Suppose the eigenvalues are real. If |λ1 | > 1 then |λ2 | = 1/|λ1 | < 1 and vice versa. So one
of the products λ1n−1 , λ2n−1 will be growing large, and one will be getting small. So the only
way that xn can be getting small is if the coefficient a1 or a2 sitting in front of the growing
product is zero.
Now let us actually compute the eigenvalues. They are
√
−E ± E 2 − 4
.
λ=
2
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IV Eigenvalues and Eigenvectors
If −2 < E < 2 then the eigenvalues are complex, so there are no bound states. The interval
[−2, 2] is the conduction band, where the electrons can move through the crystal.
If E = ±2 then there is only one eigenvalue, namely 1. In this case there actually is only
one eigenvector, so our analysis doesn’t apply. However there are no bounds states in this
case.
√
2
Now let us consider the case E < −2. Then
eigenvalue is λ1 = (−E + E − 4)/2
the large
−1
. The small eigenvalue is λ2 = (−E −
and the corresponding eigenvector is v1 =
E + λ1
√
−1
2
. We must now compute a1
E − 4)/2 and the corresponding eigenvector is v2 =
E + λ2
a
and determine when it is zero. We have [v1 |v2 ] 1 = x1 . This is 2 × 2 matrix equation
a2
a
that we can easily solve for 1 . A short calculation gives
a2
a1 = (λ1 − λ2 )−1 (−(a + E)(E + λ2 ) + 1).
Thus we see that a1 = 0 whenever
(a + E)(E −
p
E 2 − 4) − 2 = 0
Let’s consider the case a = 5 and plot this function on the interval [−10, −2]. To see if it
crosses the axis, we also plot the function zero.
>N=500;
>E=linspace(-10,-2,N);
>ONE=ones(1,N);
>plot(E,(5*ONE+E).*(E-sqrt(E.^2 - 4*ONE)) - 2*ONE)
>hold on
>plot(E,zeros(1,N))
Here is the result
156
IV.4 The Anderson Tight Binding Model
100
80
60
40
20
0
-20
-10
-9
-8
-7
-6
-5
-4
-3
-2
We can see that there is a single bound state in this interval, just below −5. In fact, the
solution is E = −5.2.
The case E > 2 is similar. This time we end up with
p
(a + E)(E + E 2 − 4) − 2 = 0
When a = 5 this never has a solution for E > 2. In fact the right side of this equation is
bigger than (5 + 2)(2 + 0) − 2 = 12 and so can never equal zero.
In conclusion, if V1 = −5, and all the other Vn ’s are zero, then there is exactly one bound
state with energy E = −5.2. Here is a diagram of the energy spectrum for this potential.
Bound state energy
−6
−4
Conduction band
−2
0
2
E
For the bound state energy of E = −5.2, the corresponding wave function Ψ, and thus the
probability that the electron is located at the nth lattice point can now also be computed.
The evaluation of the infinite sum that gives the normalization constant N 2 can be done
using a geometric series.
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IV Eigenvalues and Eigenvectors
IV.4.4 Conduction bands for a crystal
The atoms in a crystal are arranged in a periodic array. We can model a one dimensional
crystal in the tight binding model by considering potential values that repeat a fixed pattern.
Let’s focus on the case where the pattern is 1, 2, 3, 4 so that the potential values are
V1 , V2 , V3 , . . . = 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, . . .
In this case, if we start with the formula
1
xn = A(Vn − E)A(Vn−1 − E) · · · A(V1 − E)
0
we can group the matrices into groups of four. The product
T (E) = A(V4 − E)A(V3 − E)A(V2 − E)A(V1 − E)
is repeated so that
x4n
1
= T (E)
0
n
Notice that the matrix T (E) has determinant 1 since it is a product of matrices with
determinant 1. So, as above, the eigenvalues λ1 and λ2 are either real with λ2 = 1/λ1 , or
complex conjugates on the unit circle. As before, the conduction bands are the energies E
for which the eigenvalues of T (E) are complex conjugates. It turns out that this happens
exactly when
|tr(T (E))| ≤ 2
To see this, start with the characteristic polynomial for T (E)
det(λI − T (E)) = λ2 − tr(T (E))λ + det(T (E))
(see homework problem). Since det(T (E)) = 1 the eigenvalues are given by
p
tr(T (E)) ± tr(T (E))2 − 4
λ=
.
2
When |tr(T (E))| ≤ 2 the quantity under the square root sign is negative, and so the eigenvalues have a non-zero imaginary part.
Let’s use MATLAB/Octave to plot the values of tr(T (E)) as a function of E. For convenience we first define a function that computes the matrices A(z). To to this we type the
following lines into a file called A.m in our working directory.
function A=A(Z)
A=[Z -1; 1 0];
end
158
IV.4 The Anderson Tight Binding Model
Next we start with a range of E values and define another vector T that contains the corresponding values of tr(T (E)).
N=100;
E=linspace(-1,6,N);
T=[];
for e = E
T=[T trace(A(4-e)*A(3-e)*A(2-e)*A(1-e))];
end
Finally, we plot T against E. At the same time, we plot the constant functions E = 2 and
E = −2.
plot(E,T)
hold on
plot(E,2*ones(1,N));
plot(E,-2*ones(1,N));
axis([-1,6,-10,10])
On the resulting picture the energies where T (E) lies between −2 and 2 have been highlighted.
10
5
0
−5
−10
−1
0
1
2
3
4
5
6
We see that there are four conduction bands for this crystal.
159
IV Eigenvalues and Eigenvectors
IV.5 Markov Chains
Prerequisites and Learning Goals
After completing this section, you should be able to
• write down the definition of a stochastic matrix and its properties
• explain why the probabilities in a random walk approach limiting values
• write down the stochastic matrix for a random walk and calculate the limiting probabilities.
• use stochastic matrices to solve practical Markov chain problems.
• write down the stochastic matrix associated with the Google Page rank algorithm for
a given damping factor α, and compute the ranking of the sites for a specified internet.
• use the Metropolis algorithm to produce a stochastic matrix with a predetermined
limiting probability distribution
IV.5.1 Random walk
In the diagram below there are three sites labelled 1, 2 and 3. Think of a walker moving
from site to site. At each step the walker either stays at the same site, or moves to one of
the other sites according to a set of fixed probabilities. The probability of moving to the ith
site from the jth site is denoted pi,j . These numbers satisfy
0 ≤ pi,j ≤ 1
because they are probabilities (0 means “no chance” and 1 means “for sure”). On the
diagram they label the arrows indicating the relevant transitions. Since the walker has to
go somewhere at each step the sum of all the probabilities leaving a given site must be one.
Thus for every j,
X
pi,j = 1
i
160
IV.5 Markov Chains
p
p
3,3
2,2
p2,3
3
2
p3,2
p
p2,1
1,3
p
p
1,2
3,1
1
p1,1
Let xn,i be the probability that the walker is at site i after n steps. We collect these
probabilities into a sequence of vectors called state vectors. Each state vector contains the
probabilities for the nth step in the walk.
xn,1
xn,2
xn = .
..
xn,k
The probability that the walker is at site i after n+1 steps can be calculated from probabilities
for the previous step. It is the sum over all sites of the probability that the walker was at
that site, times the probability of moving from that site to the ith site. Thus
X
xn+1,i =
pi,j xn,j
j
This can be written in matrix form as
xn = P xn−1
where P = [pi,j ]. Using this relation repeatedly we find
xn = P n x0
where x0 contains the probabilities at the beginning of the walk.
The matrix P has two properties:
1. All entries of P are non-negative.
2. Each column of P sum to 1.
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IV Eigenvalues and Eigenvectors
A matrix with these properties is called a stochastic matrix.
The goal is to determine where the walker is likely to be located after many steps. In
other words, we want to find the large n behaviour of xn = P n x0 . Let’s look at an example.
Suppose there are three sites, the transition probabilities are given by
0.5 0.2 0.1
P = 0.4 0.2 0.8
0.1 0.6 0.1
and the walker starts at site 1 so that initial state vector is
1
x0 = 0
0
Now let’s use MATLAB/Octave to calculate the subsequent state vectors for n = 1, 10, 100, 1000.
>P=[.5 .2 .1; .4 .2 .8; .1 .6 .1];
>X0=[1; 0; 0];
>P*X0
ans =
0.50000
0.40000
0.10000
>P^10*X0
ans =
0.24007
0.43961
0.32032
>P^100*X0
ans =
0.24000
0.44000
0.32000
P^1000*X0
162
IV.5 Markov Chains
ans =
0.24000
0.44000
0.32000
The state vectors converge. Let’s see what happens if the initial vector is different, say with
equal probabilities of being at the second and third sites.
>X0 = [0; 0.5; 0.5];
>P^100*X0
ans =
0.24000
0.44000
0.32000
The limit is the same. Of course, we know how to compute high powers of a matrix using
the eigenvalues and eigenvectors. A little thought would lead us to suspect that P has an
eigenvalue of 1 that is largest in absolute value, and that the corresponding eigenvector is
the limiting vector, up to a multiple. Let’s check
>eig(P)
ans =
1.00000
0.35826
-0.55826
>P*[0.24000; 0.44000; 0.32000]
ans =
0.24000
0.44000
0.32000
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IV Eigenvalues and Eigenvectors
IV.5.2 Properties of stochastic matrices
The fact that the matrix P in the example above has an eigenvalue of 1 and an eigenvector
that is a state vector is no accident. Any stochastic matrix P has the following properties:
(1) If x is a state vector, so is P x.
(2) P has an eigenvalue λ1 = 1.
(3) The corresponding eigenvector v1 has all non-negative entries.
(4) The other eigenvalues λi have |λi | ≤ 1
If P or some power P k has all positive entries (that is, no zero entries) then
(3’) The eigenvector v1 has all positive entries.
(4’) The other eigenvalues λi have |λi | < 1
(Since eigenvectors are only defined up to non-zero scalar multiples, strictly speaking, (3)
and (3’) should say that after possibly multiplying v1 by −1 the entries are non-negative
or positive.) These properties explain the convergence properties of the state vectors of the
random walk. Suppose (3’) and (4’) hold and we expand the initial vector x0 in a basis of
eigenvectors. (Here we are assuming that P is diagonalizable, which is almost always true.)
Then
x0 = c1 v1 + c2 v2 + · · · + ck vk
so that
xn = P n x0 = c1 λn1 v1 + c2 λn2 v2 + · · · + ck λnk vk
Since λ1 = 1 and |λi | < 1 for i 6= 1 we find
lim xn = c1 v1
n→∞
Since each xn is a state vector, so is the limit c1 v1 . This allows us to compute c1 easily. It
is the reciprocal of the sum of the entries of v1 . In particular, if we chose v1 to be a state
vector then c1 = 1.
Now we will go through the properties above and explain why they are true
(1) P preserves state vectors:
Suppose x is a state vector, that is, x has non-negative entries which sum to 1. Then P x
has non-negative entries too, since all the entries of P are non-negative. Also
X
X
XX
XX
X X
Pi,j =
xj = 1
(P x)i =
Pi,j xj =
Pi,j xj =
xj
i
i
j
j
i
j
i
Thus the entries of P x also sum to one, and P x is a state vector.
164
j
IV.5 Markov Chains
(2) P has an eigenvalue λ1 = 1
The key point here is that P and P T have the same eigenvalues. To see this recall that
det(A) = det(AT ). This implies that
det(λI − P ) = det((λI − P )T ) = det(λI T − P T ) = det(λI − P T )
So P and P T have the same characteristic polynomial. Since the eigenvalues are the zeros of
the characteristic polynomial, they must be the same for P and P T . (Notice that this does
not say that the eigenvectors are the same.)
Since P has columns adding up to 1, P T has rows that add up to 1. This fact can be
expressed as the matrix equation
1
1
1 1
P . = . .
.. ..
1
1
But this equation says that 1 is an eigenvalue for P T . Therefore 1 is an eigenvalue for P as
well.
(4) Other eigenvalues of P have modulus ≤ 1
To show that this is true, we use the 1-norm introduced at the beginning of the course.
Recall that the norm k · k1 is defined by
x1
x2
.. = |x1 | + |x2 | + · · · + |xn |.
. 1
xn
Multiplication by P decreases the length of vectors if we use this norm to measure length.
In other words
kP xk1 ≤ kxk1
165
IV Eigenvalues and Eigenvectors
for any vector x. This follows from the calculation (almost the same as one above, and again
using the fact that columns of P are positive and sum to 1)
X
kP xk1 =
|(P x)i |
i
X X
=
|(
Pi,j xj )|
i
≤
=
=
i
j
Pi,j |xj |
XX
j
i
Pi,j |xj |
X
|xj |
Pi,j
j
=
j
XX
X
j
X
i
|xj |
= kxk1
Now suppose that λ is an eigenvalue, so that P v = λv for some non-zero v. Then
kλvk1 = kP vk1
Since kλvk1 = |λ|kvk1 and kP vk1 ≤ kvk1 this implies
|λ|kvk1 ≤ kvk1
Finally, since v is not zero, kvk1 > 0. Therefore we can divide by kvk1 to obtain
|λ| ≤ 1
(3) The eigenvector v1 (or some multiple of it) has all non-negative entries
We can give a partial proof of this, in the situation where the eigenvalues other than λ1 = 1
obey the strict inequality |λi | < 1. In this case the power method implies that starting with
any initial vector x0 the vectors P n x0 converge to a multiple c1 v1 of v1 . If we choose the
initial vector x0 to have only positive entries, then every vector in the sequence P n x0 has only
non-negative entries. This implies that the limiting vector must have non-negative entries.
(3) and (4) vs. (3’) and (4’) and P n with all positive entries
Saying that P n has all positive entries means that there is a non-zero probability of moving
between any two sites in n steps. The fact that in this case all the eigenvalues other than
λ1 = 1 obey the strict inequality |λi | < 1 follows from a famous theorem in linear algebra
called the Perron–Frobenius theorem.
Although we won’t be able to prove the Perron–Frobenius theorem, we can give some
examples to show that if the condition that P n has all positive entries for some n is violated,
then (3’) and (4’) need not hold.
166
IV.5 Markov Chains
The first example is the matrix
0 1
P =
1 0
This matrix represents a random walk with two sites that isn’t very random. Starting at site
n
one, the walker moves to site two with probability 1, and vice versa.
The powers P of P
1
are equal to I or P depending on whether n is even or odd. So P n
doesn’t converge: it
0
0
1
for odd n. The eigenvalues of P are easily computed to
for even n and
is equal to
1
0
be 1 and −1. They both have modulus one.
For the second example, consider a random walk where the sites can be divided into two
sets A and B and the probability of going from any site in B to any site in A is zero. In
this case the i, j entries P n with the i site in A and the jth site in B are always zero. Also,
applying P to any state vector drains probability from A to B without sending any back.
This means that in the limit P n x0 (that is, the eigenvector v1 ) will have zero entries for all
sites in A. For example, consider a three site random walk (the very first example above)
where there is no chance of ever going back to site 1. The matrix
1/3 0
0
P = 1/3 1/2 1/2
1/3 1/2 1/2
corresponds to such a walk. We can check
>P=[1/3 0 0;1/3 1/2 1/2; 1/3 1/2 1/2];
>[V D] = eig(P)
V =
0.00000
0.70711
0.70711
0.00000
0.70711
-0.70711
0.81650
-0.40825
-0.40825
1.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.33333
D =
0
The eigenvector corresponding to the eigenvalue 1 is 1/2 (after normalization to make it
1/2
a state vector).
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IV Eigenvalues and Eigenvectors
IV.5.3 Google PageRank
I’m going to refer you to the excellent article by David Austin:
http://www.ams.org/featurecolumn/archive/pagerank.html
Here are the main points:
The sites are web pages and connections are links between pages. The random walk is a
web surfer clicking links at random. The rank of a page is the probability that the surfer
will land on that page in the limit of infinitely many clicks.
We assume that the surfer is equally likely to click on any link on a given page. In other
words, we assign to each outgoing link a transition probability of
1/(total number of outgoing links for that page).
This rule doesn’t tell us what happens when the surfer lands on a page with no outgoing
links (so-called dangling nodes). In this case, we assume that any page on the internet is
chosen at random with equal probability.
The two rules above define a stochastic matrix
P = P1 + P2
where P1 contains the probabilities for the outgoing links and P2 contains the probabilities
for the dangling nodes. The matrix P1 is very sparse, since each web page has typically about
10 outgoing links. This translates to 10 non-zero entries (out of about 2 billion) for each
column of P1 . The matrix P2 has a non-zero column for each dangling node. The entries in
each non-zero column are all the same, and equal to 1/N where N is the total number of
sites.
If x is a state vector, then P1 x is easy to compute, because P1 is so sparse, and P2 x is
a vector with all entries equal to the total probability that the state vector x assigns to
dangling nodes, divided by N , the total number of sites.
We could try to use the matrix P to define the rank of a page, by taking the eigenvector v1
corresponding to the eigenvalue 1 of P and defining the rank of a page to be the entry in v1
corresponding to that page. There are problems with this, though. Because P has so many
zero entries, it is virtually guaranteed that P will have many eigenvalues with modulus 1 (or
very close to 1) so we can’t use the power method to compute v1 . Moreover, there probably
will also be many web pages assigned a rank of zero.
To avoid these problems we choose a number α between 0 and 1 called the damping factor.
We modify the behaviour of the surfer so that with probability α the rules corresponding
to the matrix P above are followed, and with probability 1 − α the surfer picks a page at
random. This behaviour is described by the stochastic matrix
S = (1 − α)Q + αP
168
IV.5 Markov Chains
where Q is the matrix where each entry is equal to 1/N (N is the total number of sites.) If
x is a state vector, then Qx is a vector with each entry equal to 1/N .
The value of α used in practice is about 0.85. The final ranking of a page can then be
defined to be the entry in v1 for S corresponding to that page. The matrix S has all non-zero
entries
IV.5.4 The Metropolis algorithm
So far we have concentrated on the situtation where a stochastic matrix P is given, and
we are interested in finding invariant distribution (that is, the eigenvector with eigenvalue
1). Now we want to turn the situation around. Suppose we have a state vector π and
we want to find a stochastic matrix that has this vector as its invariant distribution. The
Metropolis algorithm, named after the mathematician and physicist Nicholas Metropolis,
takes an arbitrary stochastic matrix P and modifies to produce a stochastic matrix Q that
has π as its invariant distribution.
This is useful in a situation where we have an enormous number of sites and some function
p that gives a non-negative value for each site. Suppose that there is one site where p is
very much larger than for any of the other sites, and that our goal is to find that site. In
other words, we have a discrete maximization problem. We assume that it is not difficult to
compute p for any given site, but the number of sites is too huge to simply compute p for
every site and see where it is the largest.
To solve this problem let’s assume that the sites are labeled by integers 1, . . . , N . The
vector p = [p1 , . . . , pN ]T has non-negative entries, and our goal is to find theP
largest one. We
can form a state vector π (in principle) by normalizing p, that is, π = p/ i pi . Then the
state vector π gives a very large probability to the site we want to find. Now we can use the
Metropolis algorithm to define a random walk with π as its invariant distribution. If we step
from site to site according to this random walk, chances are high that after a while we end
up at the site where π is large.
P
In practice we don’t want to compute the sum i pi in the denominator of the expression
for π vector, since the number of terms is huge. An important feature of the Metropolis
algorithm is that this computation is not required.
You can more learn about the Metropolis algorithm in the article The Markov chain Monte
Carlo revolution by Persi Diaconis at
http://www.ams.org/bull/2009-46-02/S0273-0979-08-01238-X/home.html.
In this article Diaconis presents an example where the Metropolis algorithm is used to solve a
substitution cipher, that is, a code where each letter in a message is replaced by another letter.
The “sites” in this example are all permutations of a given string of letters and punctuation.
The function p is defined by analyzing adjacent pairs of letters. Every adjacent pair of letters
has a certain probablility of occuring in an English text. Knowing these probablities is it
169
IV Eigenvalues and Eigenvectors
possible to construct a function p that is large on strings that are actually English. Here is
the output of a random walk that is attempting to maximizde this function.
IV.5.5 Description of the algorithm
To begin, notice that a square matrix P with non-negative entries is stochastic if P T 1 = 1,
where 1 = [1, 1, . . . , 1]T is the vector with entries all equal to 1. This is just another way of
saying that the columns of P add up to 1.
Next, suppose we are given a state vector π = [π1 , π2 , . . . , πn ]T . Form the diagonal matrix
Π that has these entries on the diagonal. Then, if P is a stochastic matrix, the condition
P Π = ΠP T
implies that π is the invartiant distribution for P . To see this, notice that Π1 = π so applying
the both sides of the equation to 1 yields P π = π.
This condition can be written as a collection of conditions on the components of P .
pi,j πj = πi pj,i
Notice that for diagonal entries pi,i this condition is always true. So we may make any changes
we want to the diagonal entries without affecting this condition. For the off-diagonal entries
(i 6= j) there is one equation for each pair pi,j , pj,i .
Here is how the Metropolis algorithm works. We start with a stochastic matrix P where
these equations are not neccesarily true. For each off-diagonal pair pi,j , pj,i , of matrix entries,
we make the equation above hold by by decreasing the value of the one of the entries, while
leaving the other entry alone. It is easy to see that the adjusted value will still be nonnegative.
170
IV.5 Markov Chains
Let Q denote the adjusted matrix. Then QΠ = ΠQT as required, but Q is not neccesarily
a stochastic matrix. However we have the freedom to change the diagonal entries of Q. So
set the diagonal entries of Q equal to
X
qi,i = 1 −
qj,i
j6=i
Then the columns of Q add up to 1 as required. The only thing left to check is that the
entries we just defined are non-negative. Since we have always made the adjustment so that
qi,j ≤ pi,j we have
X
qi,i ≥ 1 −
pj,i = pi,i ≥ 0
j6=i
In practice we may not be presented with state vector π but with a positive vector p =
[p1 , p2 , . . . , pn ]T whoseP
maximal component we want to find. Although in principle it is easy
to obtain π as π=p/( i pi ), in practice there may be so many terms in the sum that it is
impossible to compute. However notice that the crucial equation pi,j πj = πi pj,i is true for πi
and πj if and only if it is true for pi and pj . Thus we may use the values of p instead.
Notice that if one of pi,j or pj,i is 0 then qi,j = qj,i = 0. So it is possible that this algorithm
would yield Q = I. This is indeed a stochastic matrix where every state vector is an invariant
distribution, but it is not very useful. To avoid examples like this, one could restrict the use
of the Metropolis algorithm to matrices P where pi,j and pj,i are always either both zero or
both nonzero.
IV.5.6 An example
In this example we will use the Metropolis algorithm to try to maximize a function f (x, y)
defined in the square −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1. We put down a uniform (2N +1)×(2N +1)
grid and take the resulting lattice points to be our sites. The initial stochastic matrix P is
the one that gives equal probabilities for travelling form a site to each neighbouring site. We
use the Metropolis algorithm to modify this, obtaining a stochastic matrix Q whose invariant
distribution is proportional to the sampled values of f . We then use Q to produce a random
path that has a high probability of converging to the maximum of f .
To begin we write a MATLAB/Octave function f.m that defines a Gaussian function f
with a peak at (0, 0).
function f=f(x,y)
a = 10;
f = exp(-(a*x)^2-(a*y)^2);
end
171
IV Eigenvalues and Eigenvectors
Next we write a function p.m such that p(i,j,k,l,N) defines the matrix element for
stochastic matrix P corresponding to the sites (i, j) and (k, l) when the grid size is (2N +
1) × (2N + 1).
function p = p(i,j,k,l,N)
% [i,j] and [k,l] are two points on a grid with
% -N <= i,j,k,l <= N. The probabilities are those for
% a uniform random walk, taking into account the edges
% Note: with our convention, p([i,j],[k,l],N) is the
% probability of going from [k,l] to [i,j]
% Note: if i,j,k,l are out of range, then p=0
if( ([k,l]==[N,N] | [k,l]==[N,-N] | [k,l]==[-N,N] | [k,l]==[-N,-N])
& abs(i-k)+abs(j-l)==1)
% starting point [k,l] is a corner:
p = 1/2;
elseif( (k==N | k==-N | l==N | l==-N) & abs(i-k)+abs(j-l)==1)
% starting point [k,l] is on the edge:
p = 1/3;
elseif(abs(i-k)+abs(j-l)==1)
% starting point is inside
p = 1/4;
else
p = 0;
end
% handle out of range cases:
if(i<-N | i>N | j<-N | j>N | k<-N | k>N | l<-N | l>N )
p=0;
end
end
The next function q.m implements the Metropolis algorithm to give the matrix elements
of Q.
function qq = q(i,j,k,l,N)
if(i~=k | j~=l)
if(p(i,j,k,l,N)==0)
172
IV.5 Markov Chains
qq = 0;
elseif(p(i,j,k,l,N)*f(k/N,l/N) > f(i/N,j/N)*p(k,l,i,j,N))
qq = p(i,j,k,l,N)*f(i/N,j/N)/f(k/N,l/N);
else
qq = p(i,j,k,l,N);
end
else
qq = 1 - q(k+1,l,k,l,N) - q(k-1,l,k,l,N) - q(k,l+1,k,l,N) - q(k,l-1,k,l,N);
end
Finally, here are the commands that compute a random walk defined by Q
% set the grid size
N=50;
% set the number of iterations
niter=1000
% pick starting grid point [i,j] at random
k=discrete_rnd(1,[-N:N],ones(1,2*N+1)/(2*N+1));
l=discrete_rnd(1,[-N:N],ones(1,2*N+1)/(2*N+1));
% or start in a corner (if these are uncommented)
k=N;
l=N;
% initialize: X and Y contain the path
% F contains f values along the path
X=zeros(1,niter);
Y=zeros(1,niter);
F=zeros(1,niter);
% the main loop
for count=1:niter
% pick the direction to go in the grid,
% according to the probabilites in the stochastic matrix q
probs = [q(k,l+1,k,l,N),q(k+1,l,k,l,N),q(k,l-1,k,l,N),q(k-1,l,k,l,N),q(k,l,k,l,N)];
dn = discrete_rnd(1,[1,2,3,4,5],probs);
switch dn
case 1
l=l+1;
case 2
k=k+1;
case 3
173
IV Eigenvalues and Eigenvectors
l=l-1;
case 4
k=k-1;
end
% calculate X, Y and F values for the new grid point
X(count)=k/N;
Y(count)=l/N;
F(count)=f(k/N,l/N);
end
% plot the path
subplot(1,2,1);
plot(X,Y);
axis([-1,1,-1,1]);
axis equal
%plot the values of F along the path
subplot(1,2,2);
plot(F);
The resulting plot looks like
1
1
0.8
0.5
0.6
0
0.4
-0.5
0.2
-1
-1
-0.5
0
0.5
1
0
0
174
200
400
600
800
1000
IV.6 The Singular Value Decomposition
IV.6 The Singular Value Decomposition
Prerequisites and Learning Goals
After completing this section, you should be able to
• explain what the singular value decomposition of a matrix is
• explain why a Hermitian matrix always has real eigenvalues and an orthonormal basis
of eigenvectors.
• write down the relationship between the singular values of a matrix and its matrix
norm
IV.6.1 The matrix norm and eigenvalues
Recall that the matrix norm kDk of a diagonal matrix
λ1 0 0 · · ·
0 λ2 0 · · ·
D = 0 0 λ3 · · ·
..
..
.. . .
.
.
.
.
0
0
0
···
0
0
0
..
.
λn
is the largest absolute value of a diagonal entry, that is, the largest value of |λi |. Since,
for a diagonal matrix the diagonal entries λi are also the eigenvalues of D it is natural to
conjecture that for any matrix A the matrix norm kAk is the largest absolute value of an
eigenvalue of A.
Let’s test this conjecture on a 2 × 2 example.
A=[1 2; 3 4]
A =
1
3
2
4
eig(A)
ans =
-0.37228
5.37228
norm(A)
ans = 5.4650
175
IV Eigenvalues and Eigenvectors
Since 5.37228 6= 5.4650 we see that the conjecture is false. But before giving up on this
idea, let’s try one more time.
B=[1 3; 3 4]
B =
1
3
3
4
eig(B)
ans =
-0.85410
5.85410
norm(B)
ans = 5.8541
This time the norm and the largest absolute value of an eigenvalue are both equal to 5.8541
The difference between these two examples is that B is a Hermitian matrix (in fact, a real
symmetric matrix) while A is not. It turns out that for any Hermitian matrix (recall this
means A∗ = A) the matrix norm is equal to the largest eigenvalue in absolute value. This
is related to the fact that every Hermitian matrix A can be diagonalized by a unitary U :
A = U DU ∗ with D diagonal with the eigenvalues on the diagonal and U unitary. The point
is that multiplying A by a unitary matrix doesn’t change the norm. Thus D has the same
norm as U D which has the same norm as U DU ∗ = A. But the norm of D is the largest
eigenvalue in absolute value.
The singular value is a decompostion similar to this that is valid for any matrix. For an
arbitrary matrix it takes the form
A = U ΣV ∗
where U and V are unitary and Σ is a diagonal matrix with positive entries on the diagonal.
These positve numbers are called the singular values of A. As we will see it is the largest
singluar value that is equal to the matrix norm. The matrix A does not have to be a square
matrix. In this case, the unitary matrices U and V are not only different, but have different
sizes. The matrix Σ has the same dimensions as A.
IV.6.2 Diagonalizing Hermitian matrices
We already know that
• The eigenvalues of any Hermetian matrix are real
176
IV.6 The Singular Value Decomposition
• Eigenvectors belonging to distinct eigenvalues are orthogonal
This means that if all the eigenvalues of a Hermetian matrix A are distinct, we can choose
the correpsonding eigenvectors to form an orthonormal basis. The matrix U with with this
orthonormal basis as its columns is unitary (that is, U ∗ = U −1 ) and diagonalizes A. Thus
U ∗ AU = U −1 AU = D, where D is a diagonal matrix with eigenvalues of A as diagonal
entries.
The goal of this section is to show that any Hermitian matrix A can be diagonalized by a
unitary. In other words, there exists a unitary matrix U such that U ∗ AU = U −1 AU = D,
where D is a diagonal matrix. This equation is a compact way of saying that columns of U
form a basis of eigenvectors, with eigenvalues given by the diagonal entries of D. Since U is
unitary, this basis of eigenvectors is orthonormal.
The argument we present works whether or not A has repeated eigenvalues and also gives
a new proof of the fact that the eigenvalues are real.
To begin we show that that if A is any n × n matrix (not neccesarily Hermitian), then there
exists a unitary matrix U such that U ∗ AU is upper triangular. To see this start with any
eigenvector v of A with eigenvalue λ. (Every matrix has at least one eigenvalue.) Normalize v
so that kvk = 1. Now choose and orthonormal basis q2 , . . . , qn for the orthogonal complement
of the subspace spanned by v, so that v, q2 , . . . , qn form an orthonormal basis for all of Cn .
Form the matrix U1 with these vectors as columns. Then, using ∗ to denotes a number that
177
IV Eigenvalues and Eigenvectors
is not neccesarily 0, we have
T
v
i
qT h 1
U1∗ AU1 = . A vq2 . . . qn
..
qTn
T
v
qT h i
1
= . λvAq2 . . . Aqn
..
qTn
λkvk2 ∗ . . . ∗
λhq2 , vi ∗ . . . ∗
=
..
.. .. ..
.
. . .
λhqn , vi ∗ . . . ∗
λ ∗ ... ∗
0 ∗ . . . ∗
= . . . .
.. .. .. ..
0 ∗ ... ∗
λ ∗ ... ∗
0
= .
.
..
A2
0
Here A2 is an (n − 1) × (n − 1) matrix.
Repeating the same procedure with A2 we can construct an (n − 1)× (n − 1) unitary matrix
Ũ2 with
λ2 ∗ . . . ∗
0
Ũ2∗ A2 Ũ2 = .
.
..
A
3
0
Now we use the(n − 1) × (n − 1) unitary matrix Ũ2 to construct an n × n unitary matrix
1 0 ... 0
0
U2 = .
..
Ũ2
0
178
IV.6 The Singular Value Decomposition
Then it is not hard to see that
λ ∗ ∗ ∗ ∗
0 λ2 ∗ . . . ∗
0 0
∗ ∗
U2 U1 AU1 U2 =
.
.. ..
. .
A3
0 0
Continuing in this way, we find unitary matrices U3 , U4 , . . . , Un−1 so that
λ ∗ ∗ ∗
∗
0 λ2 ∗ . . . ∗
∗
Un−1
· · · U2∗ U1∗ AU1 U2 · · · Un−1 = 0 0
.
.. ..
.
.
. .
.
0 0
λn
Define U = U1 U2 · · · Un−1 . Then U is unitary, since the product of unitary matrices
∗
· · · U2∗ U1∗ . Thus the equation above says that U ∗ AU is upper
is unitary, and U ∗ = Un−1
triangular, that is,
λ ∗ ∗ ∗
∗
0 λ2 ∗ . . . ∗
U ∗ AU = 0 0
.. ..
..
. .
.
0 0
λn
Notice that if we take the adjoint of this equation, we get
λ 0 0 0
0
∗ λ2 0 . . . 0
∗ ∗
U A U = ∗ ∗
.. ..
..
. .
.
∗ ∗
λn
Now lets return to the case where A is Hermitian. Then A∗ = A so that the matrices
appearing in the previous two equations are equal. Thus
λ 0 0 0
0
λ ∗ ∗ ∗
∗
0 λ2 ∗ . . . ∗ ∗ λ2 0 . . . 0
0 0
∗ ∗
=
.. ..
. .
..
..
. .
.. ..
.
.
0 0
λn
∗ ∗
λn
179
IV Eigenvalues and Eigenvectors
This implies that all the entries denoted ∗ must actually be zero. This also shows that λi = λi
for every i. In other words, Hermitian matrices can be diagonalized by a unitary matrix, and
all the eigenvalues are real.
IV.6.3 The singular value decomposition
Let A be an m × n matrix. Then A can be factored as
A = U ΣV ∗
where U is an n × n unitary matrix, V is an n × n unitary matrix and Σ is an n × m diagonal
matrix with non-negative entries. (If n 6= m the diagonal of Σ starts at the top left corner
and runs into one of the sides of the matrix at the other end.) Here is an example.
√
√ √
√
1/ √3 −1/√2 −1/√ 6
3 √0 1 1
0 1
1 −1 = −1/ 3 −1/ 2 1/ 6 0
2
√
√
−1 0
0 1
0
0
1/ 3
0
2 6
The positive diagonal entries of Σ are called the singular values of A.
Here is how to construct the singular value decomposition in the special case where A is
an invertible square (n × n) matrix. In this case U Σ and V will all be n × n as well.
To begin, notice that A∗ A is Hermitian, since (A∗ A)∗ = A∗ A∗∗ = A∗ A), Moreover, the
(real) eigenvalues are all positive. To see this, suppose that A∗ Av = λv. Then, taking
the inner product of both sides with v, we obtain hv, A∗ Avi = hA∗ v, Avi = λhv, vi. The
eigenvector v is by definition not the zero vector, and because we have assumed that A is
invertible Av is not zero either. Thus λ = kAvk2 /kvk2 > 0.
Write the positive eigenvalues of A∗ A as σ12 , σ22 , . . . , σn2 and let Σ be the matrix with
σ1 , σ2 , . . . , σn on the diagonal. Notice that Σ is invertible, and Σ2 has the eigenvalues of
A∗ A on the diagonal. Since A∗ A is Hermitian, we can find a unitary matrix V such that
A∗ A = V Σ2 V ∗ . Define U = AV Σ−1 . Then U is unitary, since U ∗ U = Σ−1 V ∗ A∗ AV Σ−1 =
Σ−1 Σ2 Σ−1 = I.
With these definition
U ΣV ∗ = AV Σ−1 ΣV ∗ = AV V ∗ = A
so we have produced the singular value decomposition.
Notice that U is in fact the unitary matrix that diagonalizes AA∗ since U Σ2 U ∗ = AV Σ−1 Σ2 Σ−1 V ∗ A∗ =
AV V ∗ A∗ = AA∗ . Moreover, this formula shows that the eigenvalues of AA∗ are the same as
those of A∗ A, since they are the diagonal elements of Σ2 .
In MATLAB/Octave, the singular value decomposition is computed using [U,S,V]=svd(A).
Let’s try this on the example above:
180
IV.6 The Singular Value Decomposition
[U S V] = svd([1 1;1 -1;0 1])
U =
0.57735
-0.57735
0.57735
-0.70711
-0.70711
0.00000
-0.40825
0.40825
0.81650
S =
1.73205
0.00000
0.00000
0.00000
1.41421
0.00000
V =
0
1
-1
-0
IV.6.4 The matrix norm and singular values
The matrix norm of a matrix is the value of the largest singular value. This follows from the
fact that multiplying a matrix by a unitary matrix doesn’t change its matrix norm. So, if
A = U ΣV ∗ , then the matrix norm kAk is the same as the matrix norm kΣk. But the Σ is a
diagonal matrix with the singular values of A along the diagonal. Thus the matrix norm of
Σ is the largest singular value of A.
Actually, the Hilbert Schmidt norm p
is also related to the singular values. Recall that for
T
a matrix with real entries
p kAkHS = tr(A A). For∗ a matrix with complex entries, the
∗
definition is kAkHS = tr(A A). Now, if A = U ΣV is the singular value decomposition
of A, then A∗ A = V Σ2 V ∗ . Thus tr(A∗ A) = tr(V Σ2 V ∗ ) = tr(V ∗ V Σ2 ) = tr(Σ2 ). Here
∗
we used that
qPtr(AB) = tr(BA) for any two matrices A and B and that V V = I. Thus
2
kAkHS =
i σi , where σi are the singular values.
Notice that if we define the vector of singular values σ = [σ1 , . . . , σn ]T then kAk = kσk∞
and kAkHS = kσk2 . By taking other vector norms for σ, like the 1-norm, or the p-norm for
other values of p, we can define a whole new family of norms for matrices. These norms are
called Schatten p-norms and are useful in some situations.
IV.6.5 Applications of the SVD
The idea is that setting all but the largest singular values of matrix to zero gives an approximation of a matrix that has low rank. More explanation will have to wait for the next
version of these notes!
181
