Linear Algebra: Concepts and Methods
Solutions to Problems
Martin Anthony and Michele Harvey
Department of Mathematics
The London School of Economics and Political Science
2
Introduction
This document contains solutions to all the Problems from the book ‘Linear
Algebra: Concepts and Methods’. It is intended as an aid to those who use the
text in support of teaching a course in linear algebra.
3
4
Chapter 1
Problem 1.1 (a) Ab is not defined.
(b) Since C is a 3 × 3 matrix and A is 3 × 2, the product CA is defined and is the
3 × 2 matrix
4 6
2 1
1 2 1
CA = 3 0 −1 1 1 = 6 0
9 8
0 3
4 1 1
(c) A + Cb is not defined, since Cb is 3 × 1 and A is 3 × 2.
2
(d) A + D = 3
6
2
6
6
(e) bT D = ( 1
0
1 −1 ) 2
6
1
5 = ( −4 3 )
3
(f) The product DAT is a 3 × 3 matrix, as is C, so this expression is defined:
)
0 1 (
1 1 3
2 1 0
DAT = 2 5
= 9 7 15 ,
1 1 3
6 3
15 9 9
5
CHAPTER 1.
6
1 1 3
1 2 1
2 3
T
DA + C =
9 7 15 + 3 0 −1 = 12 7
15 9 9
4 1 1
19 10
4
14
10
1
(g) bT b = ( 1 1 −1 ) 1 = (3). As mentioned in Section 1.8.2, we
−1
normally just identify a 1 × 1 matrix by its entry. So we may write (3) as 3.
1
(h) bbT = 1 ( 1
−1
1
1 −1
1 −1 ) = 1
1 −1
−1 −1 1
1 2 1
1
2
(i) Cb = 3 0 −1
1
= 4
4 1 1
−1
4
Problem 1.2 The matrix aT b is the product of a 1 × n matrix and an n × 1
matrix, so it is a 1 × 1 matrix (which can be identified with the scalar it
represents; that is, the scalar given by the inner product ⟨a, b⟩ ). The product bT a
is also a 1 × 1 matrix. Since the product aT b is a 1 × 1 matrix, it is symmetric,
so that
aT b = (aT b)T = bT a.
Problem 1.3 We have
(
)(
3 7
x
0 −1
z
y
w
)
(
=
3x + 7z
−z
3y + 7w
−w
)
(
=
1
0
0
1
)
Hence,
3x + 7z = 1
3y + 7w = 0
−z = 0
−w = 1
=⇒
Therefore B −1 =
(1
3
0
7
3
−1
3x + 0 = 1
7
1
=⇒ x = , y = , z = 0, w = −1
3
3
3y − 7 = 0
)
.
7
Using the method in Activity 1.24, we have
|B| = −3,
so that
B
−1
1
=−
3
(
−1 −7
0
3
)
,
which, of course, agrees with the previous result for B −1 .
You can now check that this is indeed B −1 by multiplying out BB −1 and B −1 B
to obtain the identity matrix.
Problem 1.4 To obtain A, you can first take the transpose of both sides of the
equation
)
(
3 5
−1 T
(A ) =
1 2
to obtain
(
)
3 1
−1
A =
.
5 2
Now A = (A−1 )−1 , so we need the inverse of the above matrix. This can be
found using the method given in Activity 1.24. You should obtain
(
)
2 −1
A=
.
−5 3
Problem 1.5 Since A is symmetric, we have aij = aji , and since B is
skew-symmetric we must have bij = −bji . Then the matrices are
1 −4 7
0 −3 5
A = −4 6 0
B= 3
0 2.
7
0 2
−5 −2 0
Problem 1.6 Since (A + AT )T = AT + A = A + AT , the matrix A + AT is
symmetric. Similarly, (A − AT )T = AT − A = −(A − AT ), so this matrix is
skew symmetric.
Adding the matrices A + AT and A − AT we obtain 2A, so we can write
1
1
A = (A + AT ) + (A − AT ),
2
2
which is the sum of a symmetric and a skew-symmetric matrix.
CHAPTER 1.
8
(
)
(
)
a b
1 0
Problem 1.7 We have A =
. Let B =
. Then
c d
0 0
(
)(
) (
)
a b
1 0
a 0
AB =
=
,
c d
0 0
c 0
)(
) (
)
(
1 0
a b
a b
BA =
=
.
0 0
c d
0 0
Then, AB = BA implies b = 0 and c = 0.
(
)
0 1
Now take another 2 × 2 matrix, C =
. We already have b = c = 0.
0 0
Therefore,
)(
) (
)
(
0 1
0 a
a 0
=
,
AC =
0 d
0 0
0 0
(
)(
) (
)
0 1
a 0
0 d
CA =
=
.
0 0
0 d
0 0
So, AC = CA implies a = d, and so
(
)
a 0
A=
= aI, for some a ∈ R.
0 a
Note that (aI)M = aM = M (aI), so the matrix aI commutes with all 2 × 2
matrices M .
The proof extends to 3 × 3 matrices, and it is true in general that the only
matrices which commute with all other matrices of the same size are scalar
multiples of the identity matrix, λI, for λ ∈ R.
Problem 1.8 A vector equation of the line through A = (4, 5, 1) and
B = (1, 3, −2) can be given by
x = b + t(a − b), t ∈ R,
where a and b are the position vectors of the points A and B respectively. We
have a − b = (3, 2, 3)T , so
x
1
3
y = 3 + t 2 , t ∈ R.
z
−2
3
9
The point C = (c, d, −5) is on this line if there is some t ∈ R such that
c
1
3
d = 3 + t2.
−5
−2
3
Equating components, we see that
c = 1 + 3t,
d = 3 + 2t,
−5 = −2 + 3t.
From the last equation, we must have t = −1, so that c = −2 and d = 1. The
point is (−2, 1, −5).
Problem 1.9 It is clear that the equations do not represent the same line since
(1, 1, 1)T (a point on the first line) does not satisfy the Cartesian equations of the
second line. So either they intersect in at most one point, or they are skew, or
parallel. To determine which, one way to approach this question is to substitute
the components of the vector equation into the Cartesian equations. This
immediately gives t = 2, as the only solution to both equations from which you
can conclude that there is a unique point of intersection. Using the vector
equation of the first line, the point of intersection (5, 3, −1) can be obtained.
Alternatively, you can find a vector equation of the second line by setting
y−4=
so that
z−1
= t,
2
x=5
5
0
x
5
= 4 + t 1.
t+4
x= y =
2
1 + 2t
1
z
Then the lines intersect if there are numbers t and s such that
1
2
5
0
1 + t 1 = 4 + s1.
1
−1
1
2
Equating components, we find t = 2 (and s = −1), so the point of intersection is
(5, 3, −1).
CHAPTER 1.
10
Problem 1.10 The lines are not parallel since their direction vectors are not
scalar multiples of one another. Therefore, either they intersect or they are skew.
So you need to determine if there are real numbers s and t such that
x
1
3
8
1
x = y = 2 + t −2 = 4 + s −2 .
z
1
1
−3
2
Equating components we obtain the equations
1 + 3t = 8 + s,
2 − 2t = 4 − 2s,
1 + t = −3 + 2s.
The first two equations have the solution t = 4, s = 5, but substitution of these
values into the third equation shows that is it not satisfied. Therefore there are no
values of s and t which satisfy all three equations simultaneously, and the lines
are skew.
Problem 1.11 The Cartesian equation is of the form
⟨x, n⟩ = x − 4y + 2z = d,
where d is found by substituting in the point (5, 1, 3). Therefore the plane has
equation
x − 4y + 2z = 5 − 4(1) + 2(3) = 7.
To find a vector equation, solve for one of the variables in terms of the other two;
say, x = 7 + 4y − 2z. Then,
−2
4
7
7 + 4y − 2z
x
y =
= 0 + y 1 + z 0 , y, z ∈ R.
y
1
0
0
z
z
This is a vector equation of the line.
Problem 1.12 The easiest way to find the Cartesian equation of the given plane
is to first find the Cartesian equation of a parallel plane through the origin by
equating the components of the equation
x
2
0
y = s 1 + t 1 = sv + tw,
z
−1
2
11
and eliminating s and t from the three resulting equations. You should obtain
3x − 4y + 2z = 0. Check that this is correct by verifying that the normal to the
plane, n = (3, −4, 2)T is orthogonal to both of the direction vectors, v and w,
which are in the plane.
Alternatively, you can find a normal to the plane by finding a vector
n = (a, b, c)T which is orthogonal to both v and w, by solving the simultaneous
linear equations
⟨n, v⟩ = 0 and ⟨n, w⟩ = 0.
Then the Cartesian equation of the given plane is 3x − 4y + 2z = d, where d is
obtained by substituting in the point (1, 1, 1), so d = 1. The Cartesian equation
of the plane is therefore
3x − 4y + 2z = 1.
To show that the second vector equation represents the same plane, you can show
that the point (1, 2, 3) also satisfies this equation and that the non-parallel vectors
(6, 2, −5)T and (2, −2, −7)T are also orthogonal to n. (Alternatively, you could
show that there are values of s and t such that (1, 2, 3)T satisfies the first vector
equation and that the vectors (6, 2, −5)T and (2, −2, −7)T are each a linear
combination of the direction vectors v and w.)
Problem 1.13 A vector equation of the plane is
0
2
1
x = 1 + s 1 + t1,
2
−1
1
s, t ∈ R,
since it contains the direction vector of each line and a point on one of the lines.
Problem 1.14 To find the point of intersection of the line
x
2
6
x = y = 1 + t −1
z
4
2
with the plane x + y − 3z = 4, substitute the components of the line, x = 2 + 6t,
y = 1 − t, z = 4 + 2t, into the Cartesian equation of the plane and solve for t.
You should obtain t = −13. Then the point of intersection is found by
12
CHAPTER 1.
substituting this value into the equation of the line. The point is
(x, y, z) = (−76, 14, −22). (You should check that this is correct by showing
that it does, indeed, satisfy the Cartesian equation of the second plane.)
Chapter 2
Problem 2.1 Each system of equations is solved by finding the reduced row
echelon form of the augmented matrix and then using this to obtain the solution.
(a)
1 1 1 2
1 1 1
2
R3 +R1
0 2 1 0
(A|b) = 0 2 1
0 −→
0 2 0 −2
−1 1 −1 −4
1
1
1
2
1
1
1
2
1
1
1
2
1
R
3 −2R2
3 R2
2 3
0 1 0 −1
0 1 0 −1 R−→
0 2 1 0 R−→
−→
0 0 1 2
0 2 1 0
0 1 0 −1
1 1 0 0
1 0 0 1
R1 −R3
R1 −R2
0 1 0 −1 .
−→ 0 1 0 −1 −→
0 0 1 2
0 0 1 2
So
x
1
y = −1 .
2
z
(Note that R3
R2 means interchange the second and third rows.) This system
has a unique solution, and the three planes determined by these equations
intersect in one point.
13
CHAPTER 2.
14
(b)
1
(A|b) =
0
−1
1 2 2
1 1
R3 +R1
2 1 0
−→
0 2
1 −1 0
0 2
2
1
1
2
1
R3 −R2
0
−→
0
2
0
1
2
0
2
1
0
2
0
2
This system is inconsistent: there is no solution.
The equations represent three planes which have no points in common. No two
of the planes are parallel, so any two planes intersect in a line, and these three
lines of intersection are parallel.
(c)
1 1 2
2
1 1 2 2
R3 +R1
0 2 1 0
(A|b) = 0 2 1
0 −→
−1 1 −1 −2
0 2 1 0
1 0 32
1 1 2 2 1R
1 1 2 2
R1 −R2
R3 −R2
2 2
1
−→ 0 2 1 0 −→ 0 1 2 0 −→ 0 1 12
0 0 0 0
0 0 0
0 0 0 0
2
0
0
There are infinitely many solutions since there is no leading one in the third
column. Set z = t and then use the reduced equations to solve for x and y in
terms of t. We obtain
3
−2
2 − 32 t
2
x
y = − 1 t = 0 + t − 1 , t ∈ R.
2
2
0
z
t
1
These three planes intersect in a line. Alternatively, if you set z = 2s, then the
line of solutions can be written as,
x
2
−3
y = 0 + s −1 , s ∈ R.
z
0
2
(d) The first step in reducing this matrix should probably be to switch row 1 and
row 3, so that you have a leading one without introducing fractions. The
reduced row echelon form is
1 0 0 0
x
0
(A|b) −→
0 1 0 0
=⇒ y = 0
0 0 1 0
z
0
15
So x = 0 is the unique solution. Since the system is homogeneous, we could
have just reduced the coefficient matrix A. But the question said to reduce the
augmented matrix, which emphasises the point that the last column remains all
zeros throughout the row reduction.
These equations represent 3 planes, all of which pass through the origin, and
which have one point of intersection, namely the origin.
Problem 2.2 Write down the augmented matrix and put it into reduced row
echelon form:
1 1 1 0 1 1
1
1
1
0
1
1
1
R
3 2
1 1 2 1 3 2
(A|b) = 3 3 6 3 9 6 −→
2 2 4 1 6 5
2 2 4 1 6 5
1 1 1 0 1 1
1 1 1 0 1 1
R2 −R1
3 −2R2 )
−→ 0 0 1 1 2 1 (−1)(R
0 0 1 1 2 1
−→
R3 −2R1
−→
0 0 2 1 4 3
0 0 0 1 0 −1
1 1 0 0 −1 −1
1 1 1 0 1 1
R1 −R2
R2 −R3
0 0 1 0 2
2 .
−→ 0 0 1 0 2 2 −→
0 0 0 1 0 −1
0 0 0 1 0 −1
Here, x2 and x5 are non-leading variables. Set x2 = r and x5 = s. Looking at
the equations given by the reduced row echelon form, solve for the other
variables in terms of these. The general solution is
x1
−1 − r + s
−1
−1
1
x2
0
1
0
r
x3 = 2 − 2s = 2 + r 0 + s −2 , r, s ∈ R.
x4
−1
0
0
−1
x5
s
0
0
1
Problem 2.3 The system of equations is Ax = b where
x
−5 1 −3 −8
y
A= 3 2 2
5 ,
x=
z ,
1 0 1
2
w
3
b = 3 .
−1
16
CHAPTER 2.
Putting the augmented matrix into reduced row echelon form (not all steps are
shown):
−5 1 −3 −8 3
1 0 1 2 −1
(A|b) = 3 2 2
5
3 −→ · · · −→ 0 1 2 2 −2
1 0 1
2 −1
0 0 1 1 −2
1 0 0 1 1
−→ 0 1 0 0 2 .
0 0 1 1 −2
Setting the non-leading variable, w, equal to an arbitrary parameter, t, the general
solution is
x
1−t
1
−1
y 2 2
=
=
+ t 0 ,
t ∈ R.
x=
z −2 − t −2
−1
w
t
0
1
The solution of the associated homogeneous system, Ax = 0 is
x
−1
y
= t 0 ,
x=
t ∈ R.
z
−1
w
1
Problem 2.4 Using the properties of the reduced row echelon form, for example
that the first number in a non-zero row is a leading 1, the matrix must be
1 0 0 2 2
0 1 0 2 2.
0 0 1 −4 3
If C is the reduced row echelon form of the augmented matrix of a system of
linear equations, Ax = b, with x = (x1 , x2 , x3 , x4 )T , then the solution is
obtained by setting the non-leading variable x4 to be equal to an arbitrary
parameter t. Then, reading from the matrix C,
−2
2
2 − 2t
x1
−2
x2 2 − 2t 2
x=
x3 = 3 + 4t = 3 + t 4 , t ∈ R.
1
0
t
x4
17
If C is the reduced row echelon form of a matrix B, then there are two
non-leading variables. If x = (x1 , x2 , x3 , x4 , x5 )T , then the general solution of
Bx = 0 is obtained by setting x4 = s and x5 = t. Finding the solution from the
matrix C, we have
−2
−2
−2s − 2t
x1
−2
−2
x2 −2s − 2t
x=
x3 = 4s − 3t = s 4 + t −3 , s, t ∈ R.
x4
0
1
s
1
0
t
x5
Problem 2.5 For each of the systems, put the augmented matrix into reduced
row echelon form. We have
3 1 −1 11
1 0 0 2
A = 1 1 0
2 → ··· → 0 1 0 0 ,
2 1 2 −6
0 0 1 −5
so there is a unique solution, x = (2, 0, −5)T .
For the system Bx = b, we have
1 0 5 3 2 11
1
B = 0 2 4 2 2 2 → ··· → 0
−1 5 5 0 1 −6
0
0
1
0
5
2
0
with general solution
x1
11
−5
1
x2 1
−2
0
x = x3 = 0 + s 1 + t
0 ,
x4 0
0
−1
x5
0
0
1
0 −1 11
0 0
1 ,
1 1
0
s, t ∈ R.
Problem 2.6 Putting the matrix B into reduced row echelon form, we have
1 1
4
5
1 1 4 5
0 −1 −3 −5
2 1 5 5
−→ · · ·
B=
0 1 3 2 −→ 0 1
3
2
0 2
6
7
−1 1 2 2
CHAPTER 2.
18
1
0
−→
0
0
1
1
0
0
4
3
0
0
5
1
2
0
−→
0
1
0
0
1
1
0
0
4
3
0
0
0
1
0
0
−→
0
1
0
0
0
1
0
0
1
3
0
0
0
0
.
1
0
(a) The homogeneous system Bx = 0 represents four equations in four
unknowns, say x, y, z, w. The reduced row echelon form shows that there is a
non-trivial solution. Assign a parameter t to the non-leading variable z. Then the
general solution is
−1
−t
x
y −3t
= t −3 .
=
x=
1
z t
0
0
w
(b) Since the last row of the reduced row echelon form is all zeros, it is possible
to find a vector b ∈ R4 such that Bx = b is inconsistent. If you guess a random
vector for b, it is likely that the system is inconsistent. You need a vector b for
which the row echelon form of the augmented matrix (B|b) has a leading one in
the last column. One very simple choice for this is to let b = (0, 0, 0, 1)T , since
none of the row operations will alter the zeros. Whatever you choose for b, you
need to reduce the (4 × 5) augmented matrix to show that there is a leading one
in the final column, and therefore the system is inconsistent.
(c) By taking d equal to any of the columns of B, or any linear combination of
the columns of B, you can ensure that the last row of the reduced row echelon
form will remain all zeros, and the system will be consistent. If, for example,
d = (1, 2, 0, −1)T , the first column of B, then the general solution of Bx = d is
x
1
−1
y 0
= + t −3 , t ∈ R.
x=
z 0
1
w
0
0
(This solution can be deduced using the ideas embodied in Theorem 2.29 and
Theorem 1.38.)
19
Problem 2.7 The first thing to do is to rewrite the equations
4x
−
y
+
z
=
6x
−2x − y + z = 0
−x + 4y − z = 6y
−x − 2y − z = 0 ,
as
x − y + 4z = 6z
x − y − 2z = 0
then solve the resulting homogeneous system of equations. Reducing the
coefficient matrix of this homogeneous system, we have
−2 −1 1
1 0 −1
−1 −2 −1 −→ · · · −→ 0 1 1 ,
1 −1 −2
0 0 0
with solutions
x
t
1
y = −t = t −1 ,
z
t
1
t ∈ R.
Problem 2.8 To answer the question, attempt to put the augmented matrix into
reduced row echelon form:
1 0 1 a
1 0 1
a
0 1 1 b
b
→ ··· → 0 1 1
−1 0 3 c
0 0 −2 d − 3a − b .
3 1 2 d
0 0 4
c+a
At this point, you can conclude that the system will be consistent if and only if
c + a = −2(d − 3a − b). Alternatively, do one more row operation, R4 + 2R3 .
This will give a bottom row of (0, 0, 0, c + a + 2(d − 3a − b)), and the system
will be consistent if and only if c + a + 2(d − 3a − b) = 0. Simplifying either of
these expressions, it can be seen that for the system to be consistent we must have
5a + 2b − c − 2d = 0.
Since the system Bx = b must be consistent when b is taken to be a column of
B, you should certainly check that the equation you found holds when a, b, c, d
are the components of a column of B, for each of the columns.
If Bx = b is consistent, the solution is unique, because Bx = 0 has only the
trivial solution. Why? The row echelon form of B will have a leading one in
CHAPTER 2.
20
each column (since B consists of the first three columns of the augmented matrix
above), so the only solution of the associated homogeneous system is x = 0. All
solutions of Bx = b are of the form x = p + w where w ∈ N (B), and since for
this matrix B we know that N (B) = {0}, if Bx = b is consistent, then the
solution is unique.
Problem 2.9 (a) Multiplying the matrices W R you should obtain
1.05 0.95 1.0
W R = ( 5000 2000 0 ) 1.05 1.05 1.05 = ( 7350 6850
1.20 1.26 1.23
7100 ) .
Note that the total invested was 7000, and there is a loss if state 2 occurs.
(b) We have
U R = ( 600
8000
1.05 0.95
1000 ) 1.05 1.05
1.20 1.26
1.0
1.05 = ( 10230
1.23
10230
10230 ) .
Here, the total invested was 9600, and the same profit is made in each state.
(c) Z is an arbitrage portfolio because its components sum to zero (it costs
nothing) and
1.05 0.95 1.0
ZR = ( 1000 −2000 1000 ) 1.05 1.05 1.05 = ( 150 110 130 ) ,
1.20 1.26 1.23
so it never makes a loss and it makes a profit in at least one state. (In fact, it
makes a profit in all three states since Z represents a total investment of 0 and the
values in all states are positive).
For an arbitrage portfolio with a greater profit, invest in stocks rather than land.
For example, if X = ( 0 −2000 2000 ), then XR = ( 300 420 360 ).
(d) To show that state prices do not exist, you can attempt to solve Rp = u by
row reduction of the augmented matrix,
1.05 0.95 1.0 1
1.05 1.05 1.05 1
R3 R2
1.05 0.95 1.0 1
(R|u) = 1.05 1.05 1.05 1 −→
1.20 1.26 1.23 1
1.20 1.26 1.23 1
21
1
1
1
1/1.05
1
1
1
1/1.05
R2 −1.05 R1
(1/1.05)R1
−→
0 −0.10 −0.05
−→ 1.05 0.95 1.0
1 R3 −1.20
0
R1
−→
1.20 1.26 1.23
1
0 −0.06 −0.03 −1/7
1
1
1
1/1.05
1 1 1 1/1.05
−10 R2
R +0.06 R
−→ 0
1
0.5
0 3 −→ 2 0 1 0.5
0 .
0 −0.06 −0.03 −1/7
0 0 0
−1/7
The system is inconsistent, so state prices do not exist for the given matrix R.
Alternatively, performing the row operation R2 − R1 on the augmented matrix,
you obtain
1.05 0.95 1.0 1
1.05 0.95 1.0 1
R2 −R1
0
(R|u) = 1.05 1.05 1.05 1 −→
0.10 0.05 0 .
1.20 1.26 1.23 1
1.20 1.26 1.23 1
The second row yields the equation
0.10p2 + 0.05p3 = 0,
so that both p2 and p3 cannot be positive numbers. Therefore, even if a solution
to the system Rp = u did exist, it could not be a vector of state prices.
Problem 2.10 The first set of equations is
{
x +
y
=
x + 1.02y =
51.11
2.22 .
For the solution, we have,
(
)
(
)
1
1
51.11 R2 −R1 1
1
51.11
−→
1 1.02 2.22
0 0.02 −48.89
(
)
(
)
1 1
51.11
1 0 2, 495.61
50R2
R1 −R2
−→
−→
.
0 1 −2, 444.50
0 1 −2, 444.50
So the solution is
( ) (
)
x
2495.61
=
.
y
−2444.5
CHAPTER 2.
22
For the second set of equations,
{
x +
x +
we have
y
1.016y
= 51.106
= 2.218 ,
(
)
(
)
1
1
51.106 R2 −R1 1
1
51.106
−→
1 1.016 2.218
0 0.016 −48.888
(
)
(
)
1 1
51.106
1 0 3, 106.606
62.5R2
R1 −R2
−→
−→
.
0 1 −3, 055.50
0 1 −3, 055.50
So the solution is
( ) (
)
x
3106.61
=
.
y
−3055.5
Why is there such a significant difference in the solutions?
Notice that the two lines are ‘nearly parallel’ (look at their slopes), and their
point of intersection is far away from the origin. Perturbing one of the equations
to make them ‘slightly more nearly parallel’, moves this point of intersection
much further away.
Chapter 3
Problem 3.1 For the matrix A:
1 2 3
1 0 0
(A|I) = 2 3 0
0 1 0
0 1 2
0 0 1
1 2
3
1 0 0
R2 − 2R1 0 −1 −6
−2 1 0
0 1
2
0 0 1
1 2 3
1 0 0
−1R2 0 1 6
2 −1 0
0 1 2
0 0 1
1 2 3
1
0 0
R3 − R2 0 1 6
2 −1 0
0 0 −4
−2 1 1
1 2 3
1 0
0
− 14 R3 0 1 6
2 −1 0
1
0 0 1
− 14 − 41
2
3
3
1 2 0
− 12
4
4
R2 − 6R3
3
1
0 1 0
−1
2
2
R1 − 3R3
1
1
1
−
−
0 0 1
2
4
4
3
1
9
1 0 0
−4 −4
2
3
R1 − 2R2 0 1 0
.
−1 12
2
1
1
1
−4 −4
0 0 1
2
23
CHAPTER 3.
24
3
So
2
− 41
− 94
1
2
− 41
− 14
A−1 = −1
1
2
3
2
.
You should check that AA−1 = I.
For the matrix B:
1 2 3
1 0 0
(B|I) = 2 3 0
0 1 0
0 1 6
0 0 1
1 2
3
1 0 0
R2 − 2R1 0 −1 −6
−2 1 0 .
0 1
6
0 0 1
The next step will yield a row of zeros in the reduced form of B. Therefore the
matrix B is not invertible.
For C, we have
1
0
(C|I) =
0
0
R3
R4
R1 − 4R3
So we see that
0
1
0
0
4
0
0
1
0
0
1
0
1 0 4 0
0 1 0 0
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1
0
C −1 =
0
0
0
1
0
0
0
0
0
1
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
1 0 0 −4
0 1 0 0
0 0 0 1
0 0 1 0
1
0
0
0
0
1
0
0
0 −4
0 0
.
0 1
1 0
0
0
1
0
25
You should check that CC −1 = I.
C is not an elementary matrix. Since two row operations were required to reduce
it to the identity matrix, C is the product of two elementary matrices which
correspond to (the inverses of) those two operations. If
1 0 0 0
1 0 4 0
0 1 0 0
and E2 = 0 1 0 0 ,
E1 =
0 0 0 1
0 0 1 0
0 0 1 0
0 0 0 1
then C = E2 E1 . Note that the order is important: C ̸= E1 E2 .
Problem 3.2 You found the inverse of the matrix A in Problem 3.1 (and should
have already checked it). Using this to obtain the solution of each of the systems,
we have xr = A−1 br .
(a) For b1 the solution is
3
21
−4
− 41 − 94
1
2
1
7
3
x1 = −1 2
0 =
.
2
2
1
1
1
1
3
−4 −4
−4
2
In the same way, forb2 and
b3 the solutions are:
−1
(b) x2 = A−1 b2 = 1 ,
01
−4
−1
1
(c) x3 = A b3 =
.
2
− 14
Problem 3.3 The evaluation of each determinant is shown below using the
cofactor expansion along an appropriate row or column. Other choices are
possible.
(a) Using the cofactor expansion by row 2, we have
2 5
1 0
7 1
1
5
2 = −1
1
1
2
1
−2
7
1
5
= −1(4) − 2(−33) = 62.
1
CHAPTER 3.
26
(b) Expand by row 2 and then by row 2 of the resulting 3 × 3 determinant:
7 5 2 3
5 2 3
2 0 0 0
2 3
|A| =
= −2 2 0 0 = −2(−2)
= −20.
11 2 0 0
1 −1
57 1 −1
23 57 1 −1
(c) Expanding first by column 1 yields
1
3
0
0
2
2
1
1
1
1
6
1
0
2 1 0
2 1 0
2 1 0
0
= 1 1 6 5 − 3 1 6 5 = −2 1 6 5
5
1 1 1
1 1 1
1 1 1
.
1
(
)
6 5
1 5
= −2 2
−1
= −2(2(1) − 1(−4)) = −12
1 1
1 1
(d) You can either use a cofactor expansion, say by row 2,
0
1
0
0
1
0
0
0
0
0
1
0
0
1
0
= −1 0
0
0
1
0
1
0
0
0 = −1,
1
or recognise that this is an elementary matrix, namely the identity matrix with
row 1 and row 2 switched, so the determinant is equal to −1.
(e) Expanding each successive determinant by its first row:
0
0
0
0
0
1
0
0
0
0
6
3
0
0
0
1
9
4
0
0
2
0
8
2
0
3
9
7
7
9
1
0
2
0
3
= −1 0
4
0
5
1
6
0 0
= (−1)(3)(−2) 0 6
1 3
0
0
0
6
3
0
0
1
9
4
0
2
0
8
2
3
0
9
0
7 = (−1)(3)
0
7
1
9
0
0
6
3
0
1
9
4
1
0 6
= −36.
9 = (−1)(3)(−2)(1)
1 3
4
2
0
8
2
27
Problem 3.4 Expanding by row 1, we have
3 t −2
|B| = −1 5 3 = 3(5−3)−t(−1−6)−2(−1−10) = 6+7t+22 = 7t+28.
2 1 1
So |B| = 0 ⇐⇒ t = −4.
Problem 3.5 To evaluate each of these determinants, we first use row operations
and then (to save writing) expand the determinant by the first column.
(a)
1 −4
2 −7
1
2
2 −10
3 2
1 −4 3
2
1 −1 −3
1 −1 −3
5 1
0 1 −1 −3
=
= 6
3 −2 = 0 9
16
6 0
0 6
3 −2
−2 8
0
0 6 −6
14 4
0 −2 8
0
=
9 16
9 16
=6
= 6(−25) = −150.
6 −6
1 −1
(b) If you look carefully at this determinant, or if you start to use row operations
to obtain zeros under the leading one, you will find that the third row is equal to
twice the first row, and so performing the row operation R3 − 2R1 , will yield a
row of zeros. Therefore, we have
1
1
2
2
−1
4 −1 3
7 4 3
8 −2 6
0 5 5
9 0 9
0
1
8
1
0 = 0
7
2
2
−1
4 −1 3 0
7 4 3 8
0 0 0 0 = 0.
0 5 5 7
9 0 9 2
(c) To evaluate this determinant, we first take out the common factors in each of
the first two rows, then use row operations and expansion by the first column:
3 3a
2 2b
1 c
1
a
1 a a2
3a2
2
2
2b = 6 1 b b = 6 0 b − a
0 c−a
1 c c2
c2
a2
b − a b2 − a2
.
b 2 − a2 = 6
c − a c2 − a2
2
2
c −a
CHAPTER 3.
28
Now we take out common factors in each row, to see that the determinant is
6(b − a)(c − a)
1 b+a
= 6(b − a)(c − a)(c − b).
1 c+a
Problem 3.6 The matrix equation Ax = 0 will have non-trivial solutions if and
only if |A| = 0.
|A| =
2−λ
2
3
= (2 − λ)(1 − λ) − 6 = λ2 − 3λ − 4 = (λ − 4)(λ + 1).
1−λ
Therefore the system will have non-trivial solutions if and only if
λ = 4 or λ = −1.
Problem 3.7 The first thing you should do for each matrix is evaluate its
determinant to see if it is invertible.
|A| =
2 0 −3
0 3 1 = 2(6 − 4) − 3(3) = 4 − 9 = −5, so A−1 exists.
−1 4 2
Next find the cofactors:
C11 =
3
4
C21 = −
C31 =
1
=2
2
C12 = −
0 −3
= −(12)
4 2
C22 =
0 −3
=9
3 1
Then,
0 1
= −(1)
−1 2
2 −3
=1
−1 2
C32 = −
2 −3
= −(2)
0 1
C13 =
0 3
=3
−1 4
C23 = −
C33 =
2 −12 9
1
1
−2 .
A−1 = − −1
5
3
−8
6
Check that AA−1 = I.
1 0 2
|B| = 2 1 3 = 1(1 + 3) + 2(−2) = 0, so B is not invertible.
0 −1 1
2 0
= −(8)
−1 4
2 0
=6
0 3
.
29
1 2 0
|C| = 0 1 1 = 1(−1 − 1) − 2(0 − 2) = 2, so C is invertible.
2 1 −1
Find the cofactors:
C11 = −2
C12 = (−1)(−2)
C13 = −2
C21 = (−1)(−2)
C22 = −1
C23 = (−1)(−3)
C31 = 2
C32 = (−1)(1)
C33 = 1
.
−2 2
2
1
C −1 = 2 −1 −1 .
2
−2 3
1
Then
Check your answer.
Problem 3.8 In matrix form, Ax = b, the system of equations is given by
x
2
2 −1 5
1
1 −2 y = 1 .
x
−7
−3 −2 1
Using each of the three methods, you should find that the unique solution is
−1
x = 6 .
2
Problem 3.9 (a) Using Cramer’s rule for this system, we have
1 1 1
A = 2 1 −1 ,
|A| = 1(3)−1(1)+1(5) = 7 ̸= 0,
−1 2 1
Then,
x=
8 1 1
8(3) − 1(6) + 1(3)
21
1
3 1 −1 =
=
= 3,
|A|
|A|
7
3 2 1
8
b = 3.
3
CHAPTER 3.
30
1
y=
|A|
1 8 1
7
1(6) − 8(1) + 1(9)
2 3 −1 =
= = 1,
|A|
7
−1 3 1
1
z=
|A|
1 1 8
1(−3) − 1(9) + 8(5)
28
2 1 3 =
=
= 4.
|A|
7
−1 2 3
(b) For this system of equations, it is useful to remember that the effect of a
column operation on a determinant is the same as the analogous row operation
(Theorem 3.34). Indeed, if you think about the fact that |A| = |AT |, this has to be
so. We want z in the solution of Ax = b, where
a −a b
x
a+b
A = b −b a , x = y , b = 0 .
−a 2b 3
z
a−b
To evaluate |A|, begin by adding column one to column two, replacing column 2,
exactly as you would do for a row operation. This does not change the value of
the determinant:
a
|A| = b
−a
0
0
−a + 2b
b
a b
a = −(−a + 2b)
= (a − 2b)(a2 − b2 ).
b a
3
Since a ̸= 2b and a ̸= ±b, you know that |A| ̸= 0, so Cramer’s rule can be used.
To evaluate z, you need to replace the last column of A with the vector b and
then evaluate the new determinant. So you can still use the same column
operation (replacing C2 with C2 + C1 ) to simplify the determinant:
z=
=
1
|A|
a
b
−a
−a a + b
a
1
−b
0
=
b
|A|
2b a − b
−a
−(−a + 2b) a
b
|A|
0
0
−a + 2b
a+b
0
a−b
−(a − 2b)b(a + b)
b
a+b
=
=
−
.
0
(a − 2b)(a2 − b2 )
a−b
Problem 3.10 We first prove this using determinants. Since (AB)−1 exists,
|AB| ̸= 0. Since A and B are square matrices, |AB| = |A| |B| ̸= 0, therefore
|A| ̸= 0 and |B| =
̸ 0, so both A and B are invertible.
31
To prove this using Theorem 3.12, write (AB)(AB)−1 = I. By the associativity
of matrix multiplication, this says that A(B(AB)−1 ) = I, and by the theorem
this implies that A−1 exists. Multiplying in the opposite order, (AB)−1 (AB) = I
shows in the same way that B is invertible.
32
CHAPTER 3.
Chapter 4
Problem 4.1 Put the augmented matrix into reduced row echelon form:
1 −1 1 1 2
4
1 −1 1 1 2 4
R2 +R1
0 0 1 2 1 1
(A|b) = −1 1 0 1 −1 −3 R−→
3 −R1
−→
1 −1 2 3 4
7
0 0 1 2 2 3
1 −1 1 1 2 4
1 −1 1 1 0 0
R2 −R3
R3 −R2
0 0 1 2 0 −1
−→ 0 0 1 2 1 1 R−→
1 −2R3
−→
0 0 0 0 1 2
0 0 0 0 1 2
1 −1 0 −1 0 1
R1 −R2
−→ 0 0 1 2 0 −1 .
0 0 0 0 1 2
Set the non-leading variables to arbitrary constants: x2 = s, x4 = t, and solve for
the leading variables in terms of these parameters, starting with the bottom row:
x5 = 2,
x4 = t,
x3 = −1 − 2t,
x2 = s,
x1 = 1 + s + t.
In vector form,
1
1
1
1+s+t
x1
1 0
0
x2
s
= −1 − 2t = −1 +s 0 +t −2 = p+sv1 +tv2 .
x
x=
3
0 1
0
x4
t
0
0
2
2
x5
33
CHAPTER 4.
34
We can easily verify the following:
1
0
1 −1 1 1 2
4
Ap = −1 1 0 1 −1
−1 = −3 = b,
0
1 −1 2 3 4
7
2
1
1
0
1 −1 1 1 2
= 0,
0
Av1 = −1 1 0 1 −1
0
0
1 −1 2 3 4
0
1
0
1 −1 1 1 2
0
Av2 = −1 1 0 1 −1 −2 = 0 .
1
1 −1 2 3 4
0
0
Problem 4.2 We have Ax = b, where
1
A= 0
1
1 1
1
1 −2 2 ,
0 3 −1
x
y
x=
z ,
w
3
b = 1.
2
To find a general solution, reduce the augmented matrix to reduced row echelon
form:
1 1 1
1 3
1 0 3 −1 2
(A|b) = 0 1 −2 2 1 → · · · → 0 1 −2 2 1 .
1 0 3 −1 2
0 0 0
0 0
Setting the non-leading variables z = s and w = t, the solutions are
−3
2
x
1
y 1
= + s 2 + t −2 = p + sv1 + tv2 ,
s, t ∈ R.
0
1
z 0
1
0
0
w
35
(a) The null space, N (A) is {x ∈ R4 | x = sv1 + tv2 , s, t ∈ R} where v1 , v2
are the vectors given in the above general solution.
(b) If a = (a, b, c)T is in R(A), then the reduced row echelon form of the
augmented matrix (A|a) will have the same rank as the matrix A (because the
system Ax = a will have solutions). Reducing the matrix to row echelon form,
we have
1 1 1
1 a
1 1 1 1
a
.
(A|a) = 0 1 −2 2 b → · · · → 0 1 −2 2
b
1 0 3 −1 c
0 0 0 0 c−a+b
Therefore the condition is that −a + b + c = 0.
(c) The vector d = (1, 5, 3), does not satisfy the equation of part (b), so the
system Ax = d is inconsistent.
(Note that the vector (3, 1, 2)T does satisfy this equation, and that the equations
is the Cartesian equation of a plane through the origin in R3 .)
Problem 4.3 In matrix form, the system of equations is
1 1 −2
x
1
y = 1 = b.
Cx = 2 −1 2
c 0
1
z
0
To show that this system is consistent for any c ∈ R, we calculate the
determinant of C. Using the cofactor expansion by the last row,
1 1 −2
2 −1 2 = c(0) + 1(−3) = −3 ̸= 0.
c 0
1
Since this determinant is non-zero for all c ∈ R, the system is consistent and has
a unique solution for all c ∈ R. Using, say, Cramer’s rule, to find the solution
(with not all details shown here)
1 1 −2
2
1
1 −1 2 = ,
x=
|C|
3
0 0
1
CHAPTER 4.
36
1 1 −2
1
1 − 4c
y=
2 1 2 =
,
|C|
3
c 0 1
1 1 1
1
2c
z=
2 −1 1 = − .
|C|
3
c 0 0
Problem 4.4 First express the system in matrix form, as Ax = b. Taking a
different approach from the one in the previous problem, we will row reduce the
augmented matrix. (It is easiest to begin by switching the first two rows):
2 1 1 3
1 −1
2
3
1 −1 2 3 → · · · → 0 1
−1
−1 .
1 −2 λ 4
0 0 λ−3 0
So the system is always consistent.
(a) There are no values of λ for which the system is inconsistent.
(b) If λ ̸= 3, then there is a unique solution, which is independent of λ. The
easiest way to find the solution is to continue Gaussian elimination. But the
question asks to use Cramer’s rule or an inverse matrix. Using Cramer’s rule, we
have:
2 1 1
|A| = 1 −1 2 = 3(3 − λ),
1 −2 λ
2
|A2 | = 1
1
3 1 1
|A1 | = 3 −1 2 = 6(3 − λ),
4 −2 λ
3 1
3 2 = −3(3 − λ),
4 λ
2 1
|A3 | = 1 −1
1 −2
3
3 = 0,
4
so the unique solution is x = (2, −1, 0) for any λ ̸= 3.
(c) If λ = 3, there are infinitely many solutions. This time, we replace λ with 3
and continue Gaussian elimination:
1 −1 2
3
1 0 1
2
0 1 −1 −1 → 0 1 −1 −1 ,
0 0
0
0
0 0 0
0
with general solution:
x
2−t
2
−1
x = y = −1 + t = −1 + t 1 ,
z
t
0
1
t ∈ R.
37
Problem 4.5 To solve the system of equations Bx = b using Gaussian
elimination, we begin by row reducing the augmented matrix.
1 1 2 1 3
1 1
2
1
3
2 3 −1 1 2 −→ 0 1 −5 −1 −4
1 0 7 2 −2
0 −1 5
1 −5
1 1 2
1
3
−→ 0 1 −5 −1 −4 .
0 0 0
0 −9
Therefore the system is inconsistent and there is no solution.
Since there is no vector x for which Bx = b, you can conclude that the vector b
cannot be expressed as a linear combination of the columns of B.
Problem 4.6 We assume that A is an m × n matrix and that the system of linear
equations Ax = d has the following solution:
1
2
1
2
1
1
x = 0 + s1 + t
0 = p + sv1 + tv2 , s, t ∈ R,
−1
0
−1
0
0
1
and that c1 , c2 , . . . , cn denote the columns of A. Then we can conclude the
following.
(1) The number n must be 5. Since the solutions x are in R5 , we must have
n = 5 in order for the product Ax to be defined.
(2) It is not possible to determine the number m from the information given.
However, since the solutions are of the form
x = p + a1 s1 + · · · + an−r sn−r ,
we can conclude that:
(3) the rank of A equals 3. (Therefore, the number m is at least 3.)
CHAPTER 4.
38
(4) The null space N (A) is the subset of R4 consisting of all linear combinations
of v1 and v2 ; that is,
N (A) = {x | x = sv1 + tv2 , s, t ∈ R} .
(5) Using the fact that Ap = d, we can write d = c1 + 2c2 − c4 .
(6) Using either v1 or v2 (or a linear combination of these two vectors), we can
obtain a non-trivial linear combination of the columns ci which is equal to 0. For
example, using Av1 = 0, we have 2c1 + c2 + c3 = 0.
Problem 4.7 To solve this system using Gaussian elimination, write down the
augmented matrix and put it into reduced row echelon form:
3 1 5 9 −1 11
1 0 1 2 −1 5
1 0 1 2 −1 5
3 1 5 9 −1 11
(A|b) =
−2 1 0 −1 2 −8 −→ −2 1 0 −1 2 −8
1 1 3 5
0
4
1 1 3 5
0
4
1 0 1 2 −1 5
1 0 1 2 −1 5
0 1 2 3 2 −4
−→ 0 1 2 3 2 −4
−→
0 1 2 3 0
0 0 0 0 −2 6
2
0 1 2 3 1 −1
0 0 0 0 −1 3
1 0 1 2 −1 5
1 0 1 2 0 2
0 1 2 3 2 −4
0 1 2 3 0 2
−→
0 0 0 0 1 −3 −→ 0 0 0 0 1 −3
0 0 0 0 0
0
0 0 0 0 0 0
Assign parameters s and t to the non-leading variables x3 and x4 . The general
solution is
−2
−1
2
2 − s − 2t
x1
−3
−2
x2 2 − 2s − 3t 2
0 .
+
t
1
+
s
0
=
s
x = x3 =
1
0
0
x4
t
0
0
−3
−3
x5
x = p + sv1 + tv2 .
39
The matrix A has n = 5 columns and the rank of A is r = 3 since there are three
leading ones (three non-zero rows) in the reduced row echelon form. So
n − r = 2 and the solution is of the required form with a1 = s and a2 = t.
We have Ap = b so that b = 2c1 + 2c2 − 3c5 . Any values of s and t will give
another particular solution, q. For example, if s = 1 and t = 0, we obtain
q = (1, 0, 1, 0, −3), so that b = c1 + c3 − 3c5 . Check these:
−1
1
3
11
5
= 2 1 + 2 0 − 3 −1
2
1
−2
−8
1
0
4
1
and
−1
3
5
11
5 1 1
+ − 3 −1 .
=
2
−8 −2 0
0
3
1
4
40
CHAPTER 4.
Chapter 5
Problem 5.1 The set
x
y
x+y+z =0
S1 =
z
is a subspace of R3 because it is a plane through the origin. You should also be
able to prove this directly using the definition of S1 , in the same way as you did
for Activity 5.38, as follows:
The set S1 is non-empty because 0 ∈ S1 . Let u, v ∈ S1 , α ∈ R. We note that
u1
v1
u1 + v 1
αu1
u = u2 , v = v2 =⇒ u + v = u2 + v2 , αu = αu2 .
u3
v3
u3 + v 3
αu3
To show that S1 is closed under addition, you need to show that u + v satisfies
the condition to be in S1 . Substitute its components into the equation, and use the
fact that both u and v must satisfy the condition,
(u1 + v1 ) + (u2 + v2 ) + (u3 + v3 ) = (u1 + u2 + u3 ) + (v1 + v2 + v3 ) = 0 + 0 = 0,
so u + v ∈ S1 . To show that S1 is closed under scalar multiplication, do the
same for the vector αu:
(αu1 ) + (αu2 ) + (αu3 ) = α(u1 + u2 + u3 ) = α0 = 0,
41
CHAPTER 5.
42
so αu ∈ S1 . Since S1 is non-empty and is closed under addition and scalar
multiplication, it is a subspace. This subspace of R3 is the plane through the
origin with normal n = (1, 1, 1)T .
The set
x
2
2
2
x +y +z =1
S2 =
y
z
is not a subspace. To prove this you only need one counterexample. For example,
0∈
/ S2 . This set is the unit sphere (the surface of the unit ball) centred on the
origin.
x
S3 = y x = 0
z
The set
is a subspace. To prove this you need to show that it is non-empty and closed
under addition and scalar multiplication.
The set S3 is non-empty since 0 ∈ S3 .
Let u, v ∈ S3 , α ∈ R,
0
0
u = u2 , v = v 2 .
u3
v3
Then
0
u + v = u2 + v 2 ∈ S 3
u3 + v 3
and
0
αu = αu2 ∈ S3 .
αu3
Hence the set is non-empty and closed under addition and closed under scalar
multiplication, so it is a subspace of R3 . This subspace is the yz-plane (a plane
through the origin).
43
x
S4 = y xy = 0
z
is non-empty (since 0 ∈ S4 ) and it is closed under scalar multiplication, but it is
not a subspace.
The next set,
To prove this, take for example, u = (1, 0, 2)T ∈ S4 and v = (0, 1, 1)T ∈ S4 .
These vectors are in S4 since they satisfy the condition which defines S4 , but
1
0
1
u + v = 0 + 1 = 1 ∈
/ S4 .
2
1
3
The set is therefore not closed under addition, so it is not a subspace. S4 is the
union of the xz-plane and the yz-plane.
The set
0
x
S5 = y x = 0 and y = 0 = 0 z ∈ R
z
z
is a subspace. This set is the intersection of the xz-plane and the yz-plane, which
is the z-axis. This is a line through the origin and, therefore, a subspace. You
should also be able to prove this directly from the definition of S5 .
Problem 5.2 You need to determine if each of the given vectors can be
expressed as a linear combination of u and v. Each vector equation is equivalent
to a system of three equations in two unknowns. You can solve them using
Gaussian elimination, or, in these simple cases, just solve the equations directly.
For the vector a, set
2
−1
3
αu + βv = α −1 + β 1 = −2 .
1
3
4
Then, row-reducing the augmented matrix corresponding to this system,
2 −1 3
1 −1 2
(A|a) = −1 1 −2 −→ · · · −→ 0 1 −1 .
1
3
4
0 0
6
CHAPTER 5.
44
The system is inconsistent: there is no solution, so a ∈
/ Lin{u, v}.
Clearly b = 0 ∈ Lin{u, v}, for we have 0 = 0u + 0v.
To determine if c = (7, −5, −7)T is a linear combination of u and v, solve the
system
2
−1
7
α −1 + β 1 = −5 ,
1
3
−7
which is equivalent to the component equations
(1)
2α − β = 7
−α + β = −5
(2)
α + 3β = −7
(3).
Adding equations (1) and (2) gives α = 2, and substituting this into (2) yields
β = −3. Substituting both of these values into (3) shows that
α + 3β = (2) + 3(−3) = −7
is correct. Therefore,
7
−1
2
c = 2u − 3v = 2 −1 − 3 1 = −5 .
−7
3
1
Problem 5.3 You should show that
1
1
0
1
x
α2 + β 0 + γ 1 + δ1 = y
3
−1
1
0
z
has infinitely many solutions for any vector b = (x, y, z)T ∈ R3 . This will
establish that S1 spans R3 and that any vector can be written in infinitely many
ways as a linear combination of the vectors in S1 . You need to look at the
solutions of Ax = b, where
α
1 1 0 1
x
β
A= 2 0 1 1 ,
b = y .
x = ,
γ
3 −1 1 0
z
δ
45
If the row echelon form of the coefficient matrix A has three leading ones, then
there will always be infinitely many solutions: a solution will exist since a
leading one in each row of A means the system cannot be inconsistent. Then
there will be infinitely many solutions since there must be one free (non-leading)
variable.
Row reducing A (where not all steps are shown),
1 1 0 1
1 1
0
1
1 1 0
A= 2 0 1 1
−→
0 2 −1 1
−→
0 1 − 21
3 −1 1 0
0 −4 1 −3
0 0 1
1
1
2
1
The row echelon form of the matrix A has three leading ones. Therefore, as
explained, this shows that the set S1 spans R3 , and that any vector b ∈ R3 can be
written as a linear combination of the vectors in S1 in infinitely many ways.
S2 does not span R3 . Since
|B| =
1
0
−1
2 1
1 2 = 1(9 − 6) − 1(4 − 1) = 0
3 9
the reduced echelon form of B will contain a row of zeros. There will therefore
be some b ∈ R3 for which the system is inconsistent.
For example, the vector b = (0, 0, 1)T , cannot be expressed as a linear
combination of the vectors in S2 . The system Bx = b has no solution, since
1 2 1 0
1 2 1 0
1 2 1 0
0 1 2 0 −→ 0 1 2 0 −→ 0 1 2 0 .
0 5 10 1
0 0 0 1
−1 3 9 1
S3 does not span R3 . In the same way as above, you can show that the vector
b = (0, 0, 1)T , cannot be expressed as a linear combination of the vectors in S3
by showing that the system Ax = b has no solution, where A is the matrix with
these two vectors as its (only) columns.
Two vectors span a plane in R3 . You would need more than two vectors to span
R3 .
CHAPTER 5.
46
Problem 5.4 (a) Taking the determinant of the matrix,
(
)
1 1
A=
,
2 −1
we find that |A| = −3, and since this is non-zero, A−1 exists. Then, finding A−1 ,
we have
( )
(
)
)(
) (
)
(
1 −1 −1
α
2
2
−1
−1
=A
=−
=
.
β
−5
−5
3
3 −2 1
(b) Since A−1 exists, the equation b = Ax, which is equivalent to the vector
equation
(
)
( )
( )
1
b1
1
+β
,
=α
2
−1
b2
has solution x = A−1 b, for any b ∈ R2 :
(
)( ) ( 1
( )
)
1 1 1
b1
α
b1 + 13 b2
3
= 2
=
.
b2
β
b − 13 b2
3 2 −1
3 1
(c) In general, if v and w are non-zero vectors in R2 , with v = (a, c)T and
w = (b, d)T , then Lin{v, w} = R2 ⇐⇒ αv + βw = b has a solution for any
b ∈ R2 ⇐⇒ |A| ̸= 0.
So you need to show that |A| ̸= 0 if and only if v ̸= tw for any t ∈ R. This
statement is equivalent to showing that |A| = 0 if and only if v = tw for some
t ∈ R. We will show the latter. We need to prove this both ways.
First, suppose v = tw for some t ∈ R. Then a = tb and c = td, so that
|A| =
a b
tb b
=
= tbd − tbd = 0.
c d
td d
Conversely, assume that |A|=0, and initially assume that neither b nor d equals
zero. Then
c
a
a b
= = s,
= 0 =⇒ ad = bc =⇒
c d
b
d
47
say. Then, this implies a = sb and c = sd, so that
( )
( )
a
b
v=
=s
= sw.
c
d
If b = 0, then d ̸= 0 (since w ̸= 0) and |A| = ad − bc = 0 shows that a = 0 and
therefore c ̸= 0 (since v ̸= 0), and so
( )
( )
c 0
c
0
v=
=
= w.
c
d
d
d
If d = 0, an analogous argument holds.
Problem 5.5 (a) The set S1 = {f ∈ F | f (0) = 1} is not a subspace.
The usual method is to give one counterexample. For example the zero function
0 : x → 0 is not in S1 .
S2 = {f ∈ F | f (1) = 0} is a subspace. This time the zero function 0 : x → 0
satisfies the condition to be in S2 , so S2 is non-empty.
If f, g ∈ S2 , α ∈ R then since f (1) = 0 and g(1) = 0,
(f + g)(1) = f (1) + g(1) = 0 + 0 = 0, so (f + g) ∈ S2
(αf )(1) = αf (1) = α0 = 0, so αf ∈ S2 .
Therefore, S2 is a subspace.
(b) You can show that the set S3 = {f ∈ F | f is differentiable and f ′ − f = 0}
is a subspace by proving it is closed under addition and scalar multiplication.
This means showing that (f + g) and (αf ) both satisfy the equation for any
f, g ∈ S3 and α ∈ R. Since the derivative of the sum of two functions is equal to
the sum of the derivatives, the function f + g is differentiable and
(f + g)′ (x) − (f + g)(x) = (f ′ + g ′ )(x) − (f + g)(x)
= f ′ (x) + g ′ (x) − f (x) − g(x)
= (f ′ (x) − f (x)) + (g ′ (x) − g(x))
= 0 + 0 = 0.
CHAPTER 5.
48
In the same way, since (αf )′ = αf ′ , the function αf is differentiable and
(αf )′ − (αf ) = αf ′ (x) − αf (x) = α(f ′ (x) − f (x)) = α0 = 0.
This shows that S3 is a subspace of F .
Problem 5.6 M2 (R) is a vector space under the usual matrix addition and scalar
multiplication. The 2 × 2 zero matrix 0 is the zero element and the negative of A
is −A. By the properties of matrices given in Chapter 1, all the axioms of a
vector space are satisfied.
{(
The set W1 =
a
0
0
b
)
}
| a, b ∈ R
is a subspace of M2 (R).
(
)
(
)
a 0
s 0
To show this, let
∈ W1 and
∈ W1 , and α ∈ R. Then we
0 b
0 t
have closure under addition, because
(
) (
) (
)
a 0
s 0
a+b
0
+
=
∈ W1 ,
0 b
0 t
0
b+t
and we have closure under scalar multiplication because
(
) (
)
a 0
αa 0
α
=
∈ W1 .
0 b
0 αb
So, since W1 is certainly non-empty, it is a subspace.
{(
a 1
The set W2 =
1 b
contain the zero matrix.
)
}
| a, b ∈ R
is not a subspace because it does not
{( 2
)
}
a
0
The set W3 =
| a, b ∈ R is not a subspace of M2 (R), since
0 b2
(
)
(
)
1 0
1 0
∈ W3 , but (−1)
∈
/ W3
0 1
0 1
(Note that W3 is a subspace of M2 (C).)
Chapter 6
{( ) ( )}
1
1
L1 =
,
2
3
Problem 6.1 The set
is linearly independent, since
1
2
1
̸= 0.
3
Alternatively, this can be deduced from the fact that the set L1 consists of two
vectors, and neither is a scalar multiple of the other.
The set
{( ) ( ) ( )}
1
1
4
L2 =
,
,
2
3
5
is a set of three vectors in R2 , so it is linearly dependent.
The set
{( ) ( )}
0
1
L3 =
,
0
2
is linearly dependent since the set contains the zero vector, 0.
The set
2
3
1
L4 = 2 , 7 , 5
0
0
0
49
CHAPTER 6.
50
is linearly dependent, since
1
2
0
2
7
0
3
5 = 0.
0
3
4
1
2 ,
0
, 1
L5 =
0
−1
2
is linearly independent, since
The set
1 3
2 0
0 −1
4
1 = 19 ̸= 0.
2
Problem 6.2 The set S1 is linearly independent since it consists of two non-zero
vectors and neither is a scalar multiple of the other.
To test S2 and S3 you can either write the vectors as the columns of a matrix A
and reduce it to find the number of leading ones (to determine if Ax = 0 has
non-trivial solutions), or you can see if the third vector is a linear combination of
the first two, since the first two vectors are linearly independent. We will
demonstrate both approaches.
For S2 , the vector equation
1
2
1
0
2
0
1 0
a
1 + b −1 + c 1 = 0
3
2
2
0
is equivalent to the matrix equation Ax = 0, where A is the matrix whose
columns are the given vectors. We solve this homogeneous system by putting A
into row echelon form.
1 0 0
1 2 1
0 1 0
2 0 1
A=
1 −1 1 −→ · · · −→ 0 0 1 .
0 0 0
3 2 2
51
Since there is a leading one in every column, the only solution is the trivial
solution, x = 0, so the vectors are linearly independent.
For S3 :
4
2
1
0 4
2
s
1 + t −1 = 1
8
2
3
⇐⇒
s + 2t = 4
2s = 4
s−t=1
3s + 2t = 8
From the second equation, s = 2. Substitution into the third equation gives
t = 1. These values satisfy the other two equations, so the set is linearly
dependent.
The set
1
2
4
1
2
0
4
1
S4 =
1 −1 1 1
3
2
8
2
is linearly dependent. Since S3 ⊂ S4 and we already know that S3 is linearly
dependent, the set S4 is also linearly dependent. It is still true that the third vector
is a linear combination of the first two.
Problem 6.3 The set is linearly dependent because it consists of five vectors in
R3 . Considering the question as a whole, the easiest way to answer it is to write
the vectors as the columns of a matrix A and put the matrix into reduced row
echelon form.
1 2 4 1 3
1 0 2 0 4
A = 2 0 4 1 5 → ··· → 0 1 1 0 1 .
1 −1 1 1 0
0 0 0 1 −3
From the positions of the leading ones, we can deduce that the set
W = {v1 , v2 , v4 } is a largest set of linearly independent vectors from the set S.
By finding the solutions of Ax = 0, or by looking at the columns with the
non-leading variables, you can find the dependency relations. For example,
CHAPTER 6.
52
column 3 of the reduced row echelon form tells us that v3 = 2v1 + v2 , and the
last column indicates that v5 = 4v1 + v2 − 3v4 . (You should check that these are
correct by substituting in the vectors.) Alternatively, the general solution of
Ax = 0 is
−2
−4
−1
−1
= sw1 + tw2 , s, t ∈ R.
x = s 1 + t
0
0
3
0
1
Then writing Aw1 = 0 and Aw2 = 0 as a linear combination of the columns of
A will produce the same results.
Problem 6.4 A set of vectors {v1 , v2 , . . . , vn } is linearly dependent if and only
if the vector equation
a1 v1 + a2 v2 + · · · + an vn = 0
has a non-trivial solution; that is, a solution where not all ai = 0.
The best way to determine if the given vectors are linearly dependent is to write
the vectors as the columns of a matrix A and put the matrix into reduced row
echelon form. This will also allow you to find the dependency relations between
the vectors. Reducing the matrix A, first to row echelon form and then to reduced
row echelon form, we obtain
1 1 4 5
1 1 4 5
1 0 1 0
2 1 5 5
→ ··· → 0 1 3 5 → ··· → 0 1 3 0.
A=
0 1 3 2
0 0 0 1
0 0 0 1
−1 1 2 2
0 0 0 0
0 0 0 0
The row echelon form does not have a leading one in every column; there is one
non-leading variable in the third column. Therefore the matrix equation Ax = 0
has infinitely many solutions, and the vectors are linearly dependent. This
conclusion uses the fact that if x = (x1 , x2 , x3 , x4 )T , then
Ax = x1 v1 + x2 v2 + x1 v3 + x2 v4 ,
where v1 , v2 , v3 , v4 are the columns of the matrix A.
53
You can deduce from the row echelon form that the vectors v1 , v2 , v4 are linearly
independent, and that the vectors v1 , v2 , v3 are linearly dependent. Why? The
vectors v1 , v2 , v4 correspond to the columns with the leading ones. So an
equation of the form a1 v1 + a2 v2 + a4 v4 = 0 would only have the trivial
solution. On the other hand, a vector equation of the form
b1 v1 + b2 v2 + b3 v3 = 0 would have infinitely many solutions, since the
coefficient matrix would consist of the first three columns of A, and its reduced
row echelon form would have a non-leading variable in the third column.
You can also deduce from the reduced row echelon form that v4 cannot be
expressed as a linear combination of the other vectors. The vector equation
v4 = x1 v1 + x2 v2 + x3 v3
would correspond to a system of linear equations for which A was the augmented
matrix, and the row echelon form shows that this system would be inconsistent.
Furthermore, again looking at the reduced row echelon form of A, you can see
that column three can be written as column one plus three times column two.
This linear dependency relation also holds for the columns of A, therefore,
v3 = v1 + 3v2 . This is easily checked:
4
1
1
5 2
1
=
3 0 + 31.
2
−1
1
Alternatively, you can find the linear dependency relation by finding the solution
of Ax = 0. Looking at the reduced row echelon form of A, and setting the
non-leading variable x3 equal to an arbitrary parameter t, the solution is
x1
−t
−1
x2 −3t
=
= t −3 , t ∈ R.
x=
x3 t
1
x4
0
0
Using Ax = x1 v1 + x2 v2 + x3 v3 + x4 v4 and t = 1, then Ax = 0 implies that
−v1 − 3v2 + v3 = 0,
or v3 = v1 + 3v2 ,
producing the same linear combination as before.
CHAPTER 6.
54
Problem 6.5 The set S1 spans R2 since the vectors are linearly independent
(they are not scalar multiples of one another) and there are two of them, so S1 is
a basis of R2 . Alternatively, find the determinant of the matrix with these vectors
as its columns:
1 2
̸= 0,
2 3
which shows that S1 is a basis of R2 . Another basis of Lin(S1 ) is {e1 , e2 }.
Lin(S1 ) = R2 has dimension 2.
A basis of Lin(S2 ) is {v} where v = (1, −1)T since all other vectors are scalar
multiples of it. The set Lin(S1 ) is the line
)
(
1
x=t
−1
with Cartesian equation y = −x. This is a one-dimensional subspace of R2 .
To find a basis of Lin(S3 ) or Lin(S4 ), write the vectors as the columns of a matrix
and then reduce the matrix to row echelon form. The columns corresponding to
the columns with leading ones in the echelon form will be a basis. For S3 ,
1 2 1
1 2 1
0 1 2 −→ · · · −→ 0 1 2 .
0 0 0
−1 3 9
Then a basis is given by the first two vectors. The set S3 spans a plane (through
the origin) in R3 , and Lin(S3 ) has dimension 2.
A Cartesian equation of the plane is found by observing that
1 2
0 1
−1 3
x
y =0
z
⇐⇒
x − 5y + z = 0
(You should check that all three vectors in S3 satisfy this equation.)
For S4 , write the vectors as the columns of a matrix B and reduce B to echelon
form. Columns 1, 2 and 4 have leading ones, so the first, second and fourth
55
vectors form a basis of Lin(S4 ), which is therefore a 3-dimensional subspace of
R4 . A basis of Lin(S4 ) is
1
2
1
2
0
1
,
1 −1 , 1 .
3
2
2
This is a hyperplane. Although the question only requires you to find the
Cartesian equations of lines and planes (and not hyperplanes), you can find a
Cartesian equation of the hyperplane by evaluating the determinant of a 4 × 4
matrix in a way analogous to the approach taken for S3 . This task is made far
easier by first row reducing the matrix B T to find a basis for Lin(S4 ) with 0s and
1s in the vectors.
Explicitly, since
1 2
BT = 2 0
1 1
a basis of Lin(S4 ) is
1
−1
1
1
3
2 → ··· → 0
0
2
0
1
0
1
0
1
1
1,
0
1
0
0
0
1
, ,0 .
1 0 1
1
1
0
It follows that the Cartesian equation is
1
0
1
1
0 0 x
1 0 y
= 0.
0 1 z
1 0 w
Expanding the determinant by column three, and then evaluating the resulting
3 × 3 determinant, you should find that a Cartesian equation of the hyperplane is
−x − y + w = 0.
Problem 6.6 The set S1 is not a basis. No set of four vectors in R3 can be
linearly independent.
56
CHAPTER 6.
The set S2 is not a basis. Two vectors cannot span R3 : there must be at least
three.
The set S3 is not a basis. Either notice that the third vector is the sum of the first
two, or write the vectors as the columns of a matrix A and either show that the
row echelon form does not have three leading ones or that |A| = 0. The set is
therefore not linearly independent.
The set S4 is a basis. Write the vectors as the columns of a matrix A and compute
|A| = −1 ̸= 0, which shows that S4 is a basis.
Problem 6.7 To find coordinates of w = (1, 2, −1)T with respect to each of the
bases you need to express w as a linear combination of the basis vectors.
For the standard basis, B1 , we have
0
1
0
1
1
2 = 1 0 + 2 1 − 1 0 , so [w]B1 = 2
1
−1 B1
0
0
−1
For B2 , you should first check that it is a basis of R3 using the determinant, or
start to solve the system and notice that you will get a unique solution if and only
if B2 is a basis.
To find the unique solution of
2
1
1
1
α 1 + β −1 + γ −3 = 2 ,
−1
0
−3
1
reduce the augmented matrix to reduced row echelon form. Let A be the matrix
with the given vectors as its columns. Then
1 1
2
1
1 0 0 2
(A|b) = 1 −1 −3 2 −→ · · · −→ 0 1 0 −3 ,
1 0 −3 −1
0 0 1 1
with the unique solution α = 2, β = −3, γ = 1. So the coordinates of w in the
57
basis B2 are given by
2
[w]B2 = −3
1 B2
(which you can easily check). Notice that the reduced row echelon form of the
matrix A corresponds to the first three columns of the reduced row echelon form
of the augmented matrix, and since this is the identity matrix, B2 is a basis of R3 .
Problem 6.8 Since you know that a plane has dimension two, you just need to
find two linearly independent vectors in the plane and they will be a basis. For
example, if v = (2, 1, 0)T and w = (1, 0, −1)T , then {v, w} is a basis of the
plane. Alternatively, solve the equation for, say x, to obtain x = 2y − z. Then the
plane consists of all vectors x = (x, y, z)T such that
−1
2
2y − z
x
x = y = y = y 1 + z 0 = yv − zw, y, z ∈ R.
1
0
z
z
Then {v, w} spans the plane, and is linearly independent, so it is a basis of the
plane x − 2y + z = 0.
The yz-plane is the set of all vectors of the form (0, y, z)T . So the set of vectors
{e2 , e3 } is a basis of the yz-plane.
Problem 6.9 To show
2t
H= t : t∈R
3t
is a subspace of R3 , we show it is non-empty and closed under addition and
scalar multiplication. (Also, it is a subspace because it is the linear span of the
vector v = (2, 1, 3)T ).)
Since 0 ∈ H, H is non-empty.
CHAPTER 6.
58
To show that H is closed under addition, let u, v ∈ H. Then,
2t
2s
u = t , v = s ,
3t
3s
for some s, t ∈ R. Then,
2t
2s
2t + 2s
2(t + s)
u + v = t + s = t + s = t + s ∈ H,
3t
3s
3t + 3s
3(t + s)
since (t + s) ∈ R.
To show H is closed under scalar multiplication, let u ∈ H, α ∈ R. Then
2(αt)
α(2t)
2t
αu = α t = αt = αt ∈ H,
3(αt)
α(3t)
3t
since (αt) ∈ R.
This proves that H is a subspace.
Let w ∈ H, so w = (2s, s, 3s)T for some constant s ∈ R. We need to find
constants a, b such that
0
1
2s
s = a 0 + b1,
−1
5
3s
which is equivalent to the simultaneous equations
2s = a,
s = b,
3s = −a + 5b.
Substituting the values of a and b obtained from the first two equations into the
third equation, we find that these values also satisfy the third equation. Therefore
the system has the unique solution a = 2s, b = s, and
w = (2s)v1 + (s)v2 .
59
The set of vectors {v1 , v2 } is not a basis of the subspace H since v1 ∈
/ H (and
also, v2 ∈
/ H).
2
A basis of H is {v} where v = 1 . It follows that dim(H) = 1.
3
The set {v1 , v2 } is a basis of G. It spans G since, by definition, G is the set of all
linear combinations of v1 and v2 . It is linearly independent, as neither vector is a
scalar multiple of the other.
The set H consists of the position vectors of points on a line through the origin.
The set G consists of the position vectors of points on a plane through the origin.
The line H is contained in the plane G. (H is a subspace of G.)
Problem 6.10 For the first set of equations, Ax = b1 , we have
1 1 1 1 4
1 1
1
1
4
(A|b1 ) = 2 0 1 −1 2 −→ 0 −2 −1 −3 −6
0 2 1 3 6
0 2
1
3
6
1 1
−→
0 1
0 0
1
1
1
2
3
2
0
0
1 0
4
3 −→
0 1
0
0 0
1
2
1
2
− 12
0
0
3
2
1
3.
0
Set x3 = s and x4 = t. Then
1
1
x1
1 − 21 s + 12 t
1
−2
2
x 3 − 1s − 3t 3
−3
−1
2
2
2
x= =
= + s 2 + t 2 ,
x3
0
0
1
s
x4
t
0
0
1
1
−1
1
3
+α1 −1 +α2 −3 = p+α1 s1 +α2 s2 , α1 , α2 ∈ R.
or
x=
0
2
0
2
0
0
CHAPTER 6.
60
For the second set of equations, Bx = d1 , we have
1 2 −1 −1 3
1 2 −1 −1 3
(B|d1 ) = 1 −1 −2 −1 1 −→ 0 −3 −1 0 −2
2 1 −1 0 3
0 −3 1
2 −3
5
1 2 −1 −1 3
1 0 0 23
6
2
5
0 1 0 −1
−→ 0 1 13
0
−→
·
·
·
−→
3
3
6
0 0 2
2 −1
0 0 1 1 − 21
Set x4 = α. Then
5 2
5
2
− 3α
−3
6
6
5 + 1α
5
1
x = 6 1 3 = 6 1 + α 3 = p + α1 s1 , α1 ∈ R.
−2 − α
−2
−1
α
0
1
The set of all b for which Ax = b is consistent is equal to the range of A, R(A),
which is spanned by the columns of A. Looking at the reduced row echelon form
of the augmented matrix, a basis of this space consists of the first two columns of
A, so this a two dimensional subspace of R3 , which is a plane. A Cartesian
equation of the plane is given by
1 1 x
2 0 y = 4x − 2y − 2z = 0.
0 2 z
So an equation of the plane is 2x − y − z = 0, as you can easily check by
showing that the equation is satisfied by all the columns of A. (Note that
(4, 2, 6)T is in this plane.)
Since the reduced row echelon form of the matrix B has a leading one in each
row, the set of all d such that Bx = d is consistent is the whole of R3 .
Problem 6.11 To answer these questions, first put A into reduced row echelon
form,
1 0 3 7 0
1 −2 1 1
2
−1 3 0 2 −2
−→ · · · −→ 0 1 1 3 0 .
A=
0
0 0 0 0 1
1 1 3
4
0 0 0 0 0
1
2 5 13 5
61
A basis for the range consists of the columns of A which correspond to the
leading ones in the reduced echelon form, so a basis of the range is
2
1
−2
3 −2
−1
0 , 1 , 4 .
5
2
1
A basis of the row space consists of the first three rows of the reduced row
echelon form of A (when written as vectors), so a basis of RS(A) is
0
0
1
0 1 0
3,1,0 .
7 3 0
1
0
0
A basis of the null space is obtained by solving Ax = 0. Set the non-leading
variables to arbitrary parameters, x3 = s, x4 = t. Then, reading from the reduced
row echelon form, the solutions are
x1
−3s − 7t
−3
−7
x2 −s − 3t
−1
−3
=
= s 1 + t 0 = sv1 + tv2 , s, t ∈ R.
x=
x
s
3
x4
0
1
t
x5
0
0
0
Therefore a basis of the null space is {v1 , v2 }. The null space has dimension 2.
The range has dimension 3. The rank-nullity theorem states that
dim(R(A)) + dim(N (A)) = n
where n is the number of columns of A. Since A is a 4 × 5 matrix, n = 5 and
since 2 + 3 = 5, the theorem is verified.
Problem 6.12 Reduce the matrix B to reduced echelon form,
1 2 1 3 0
1 2 1 3
0
1 2 1 3
0
B = 0 1 1 1 −1 −→ 0 1 1 1 −1 −→ 0 1 1 1 −1
1 3 2 0 1
0 1 1 −3 1
0 0 0 −4 2
CHAPTER 6.
62
(
−→
1 2
0 1
0 0
1
1
0
3 0
1 −1
1 − 21
)
1 2
0 1
−→
0 0
1
1
0
0 32
1 0
0 − 12 −→ 0 1
1 − 12
0 0
−1 0 52
1 0 − 12 .
0 1 − 12
The leading ones are in the first, second and fourth columns.
A basis of the row space is:
A basis of the range of B is:
0
0
1
1 0
0
−1 , 1 , 0 ⊂ R5 .
0 0 1
5
1
1
−
−
2
2
2
2
3
1
0 , 1 , 1 ⊂ R3 .
1
3
0
So the range of B is all of R3 .
From the reduced echelon form of B, we can find the solution of Bx = 0.
Setting the non-leading variables x3 = s and x5 = t, we have
s − 25 t
x1
1
−5
x2 −s + 12 t
−1
1
=
α
1
+
β
x3 =
0 = αx1 + βx2 , α, β ∈ R.
s
x 1t
0
1
4
2
0
2
x5
t
(Here, α is s and β is t/2.) The set {x1 , x2 } ⊂ R5 is a basis of the null space.
Therefore rank(B) = 3 and nullity(B) = 2. The number of columns is n = 5,
so that
rank(B) + nullity(B) = 3 + 2 = 5 = n.
If b = c1 + c5 , then p = ( 1 0 0 0 1 )T is a solution of Bx = b. (Why?)
Combining this with the solution of the system Bx = 0, a general solution of
63
Bx = b is
1
1
−5
0
−1
1
+ α 1 + β 0 = p + αx1 + βx2 ,
x=
0
0
0
1
1
0
2
α, β ∈ R.
Problem 6.13 First put the matrix A in reduced row echelon form,
1 0
1
1 0 1
A = 1 1
2 −→ · · · −→ 0 1 1 .
0 −1 −1
0 0 0
Therefore, rank(A) = 2.
A basis of the row space of A consists of the rows with the leading ones; a basis
of the column space consists of the two columns of the original matrix which
correspond to the columns with leading ones:
0
0
1
1
basis of RS(A) : 0 1 , basis of CS(A) : 1 1 .
−1
0
1
1
Since dim(RS(A)) = dim(CS(A)) = 2, and both are subspaces of R3 , they are
both planes. We now find their Cartesian equations.
RS(A) :
1 0 x
0 1 y = −x − y + z = 0,
1 1 z
1 0 x
1 1 y = −x + y + z = 0.
0 −1 z
These planes have different equations, and therefore they are different subspaces
of R3 .
CS(A) :
To find the null space, look at the reduced row echelon form of A, and set the
non-leading variable z to be t. We have
x
−t
−1
y = −t = t −1 ,
z
t
1
CHAPTER 6.
64
so a basis of N (A) is
−1
−1 .
1
Then dim(N (A)) = 1. The rank-nullity theorem is:
rank(A) + nullity(A) = 2 + 1 = 3 = n,
where n is the number of columns of A.
To show that the basis vector of the null space is orthogonal to the basis vectors
of RS(A), we show that the inner products are zero:
⟩
⟩
⟨
⟨
1
−1
0
−1
0 , −1 = −1+0+1 = 0,
1 , −1 = 0−1+1 = 0.
1
1
1
1
Note that the basis vector of the null space, n = (−1, −1, 1)T is a normal to the
plane RS(A).
Since the range of A is the column space of A, it is the plane −x + y + z = 0.
The system Ax = b1 is consistent if the components of b1 satisfy this equation,
and the same is true for b2 . Substituting the components into the equation, we
find that b1 ∈ R(A), but b2 ∈
/ R(A).
You can find the solution of Ax = b1 using Gaussian elimination, or by
inspection, by noticing that b1 = c1 − 2c2 , so that a general solution is
x
1
−1
x = y = −2 + t −1 , t ∈ R.
z
0
1
Problem 6.14 A basis of the row space of A is given by the first two rows of the
reduced row echelon form, {v1 , v2 }, where v1 = (1, 0, −1, 5)T and
v2 = (0, 1, 3, 2)T .
A basis of the range (the column space) consists of the columns of A
corresponding to the leading ones, so a basis is {c1 , c2 }, where c1 = (1, 2, 3)T
and c2 = (4, −1, 2)T .
65
A basis of the null space is obtained by reading the solutions of Ax = 0 from the
reduced row echelon form. Doing this, you should find that a basis of N (A) is
{s1 , s2 }, where s1 = (1, −3, 1, 0)T and s2 = (−5, −2, 0, 1)T .
The matrix equation Ax = b represents a system of three equations in four
unknowns. There are several ways to find the value of a such that Ax = b is
consistent. To be consistent, you need b ∈ R(A), so one way is to use the basis
you found for R(A) and find a solution s, t to the component equations given by
9
1
4
0 = s 2 + t −1 .
a
3
2
Alternatively, find a such that the vectors c1 , c2 , b are linearly dependent using
the determinant:
1 4 9
2 −1 0 = 9(7) + a(−9) = 0.
3 2 a
Either method yields a = 7.
The missing columns of A can be found by looking at the linear dependence
relations given by the solutions of Ax = 0. Using the basis of the null space,
As1 = c1 − 3c2 + c3 = 0 and As2 = −5c1 − 2c2 + c4 = 0, so that
13
4
1
11
4
1
c3 = − 2 + 3 −1 = −5 , c4 = 5 2 + 2 −1 = 8 .
19
2
3
3
2
3
You can check the result by row reducing the matrix A with these columns in
place.
Problem 6.15 The matrix B is 3 × 4. Using the rank-nullity theorem with n = 4,
rank(B) + nullity(B) = n = 4,
and since dim(N (B)) = 1, we must have rank(B) = 3.
The dimension of R(B) is equal to the rank of B. Since the matrix is 3 × 4, the
column space of B, which is equal to R(B), is a three dimensional subspace of
R3 . Therefore R(B) = R3 .
CHAPTER 6.
66
Since the row space of a matrix is orthogonal to the null space, the row space is
the hyperplane in R4 with Cartesian equation
x + 2y + 3z + 4w = 0,
where x = (x, y, z, w)T ∈ R4 . The vector (4, 0, 0, −1) satisfies this equation, so
it is in the row space. Since dim(RS(B)) = rank(B) = 3, you need two other
linearly independent vectors which satisfy the equation of the hyperplane. For
example, the set
3
2
4
0 −1 0
,
,
0 0 −1
−1
0
0
is linearly independent, since if
4
2
3
0
0
−1
0 0
a
0 + b 0 + c −1 = 0 ,
−1
0
0
0
then equating components shows that the only solution is b = c = a = 0. As any
set of 3 linearly independent vectors in a 3-dimensional subspace is a basis, this
set of three vectors is a basis of the row space of B.
Chapter 7
Problem 7.1 To sketch the effect of T on the standard basis, mark the unit
vector in the x direction, e1 , in one colour, and the unit vector in the y direction,
e2 in another colour (or differentiate between them by single and double
arrowheads). Then draw the other two sides of the unit square. Now draw the
vector images T (e1 ) and T (e2 ), which are given by the columns of the matrix
AT , in the same colours. Complete the image of the unit square with these
vectors as its two corresponding sides.
The linear transformation T is rotation anticlockwise by π4 radians.
Sketch the unit square and its image under S in the same way. The linear
transformation S is reflection in the y axis.
The linear transformation ST means first do T and then do S, since
ST (v) = S(T (v)). Think about what this means geometrically. First rotate the
unit square by π4 and then reflect this in the y axis. The linear transformation T S
is very different: first reflect in the y axis, and then rotate by π4 .
For the illustrations of ST and T S using the unit square, continue as above.
Draw a unit square marking e1 and e2 distinctively on it. For ST , first find the
images under T , T (e1 ) and T (e2 ) and mark these, then find the images of these
vectors under S. Now complete the unit square. When you have completed both
sketches, you will find the images under ST and T S in very different positions.
67
CHAPTER 7.
68
Calculating their matrices will verify that ST ̸= T S:
) ( 1
)
(
)( 1
√
− √12
− √2 √12
−1 0
2
AST = AS AT =
=
√1
√1
√1
√1
0 1
2
2
2
2
( 1
)(
) ( 1
)
1
√
− √2
− √2 − √12
−1 0
2
=
AT S = AT AS =
√1
√1
√1
0 1
− √12
2
2
2
The columns of AST should be ST (e1 ) and ST (e2 ). Check that this matches
your sketch. Do the same for T S.
b is a basis of R3 , since the
Problem 7.2 (a) Each of the sets B and B
determinants of the matrices with these vectors as columns are nonzero:
1
0
1
1
1
3
0
0 = 1 ̸= 0,
1
1 1
0
1 −1 1 = 3 ̸= 0.
1 0 −1
(b) The linear transformation T given by T (e1 ) = v1 , T (e2 ) = v2 , T (e3 ) = v3 ,
has matrix
1 1 0
AT = 0 1 0 ,
1 3 1
so the linear transformation is
1 1
x
T y = 0 1
1 3
z
x+y
x
0
∈ R3 .
y
0y =
1
z
x + 3y + z
(c) If S is given by: S(v1 ) = e1 , S(v2 ) = e2 , S(v3 ) = e3 , then ST = I, so
S = T −1 as linear transformations and AS = (AT )−1 as matrices.
It is straightforward to compute the inverse matrix using row operations or the
cofactor method (|AT | = 1). You should obtain
1 −1 0
AS = 0
1 0,
−1 −2 1
and then check that AT AS = I.
69
(d) The linear transformation R given by R(e1 ) = w1 , R(e2 ) = w2 ,
R(e3 ) = w3 , has matrix,
1 1
0
AR = 1 −1 1 .
1 0 −1
(e) The composite linear transformation RS is defined, since each of R and S
maps R3 to R3 . Its effect on v1 , v2 , v3 can be seen as follows:
S
R
v1 −→e1 −→w1 ,
The matrix is
S
R
v2 −→e2 −→w2 ,
S
R
v3 −→e3 −→w3 .
1 1
0
1 −1 0
1 0
0
1 0 = 0 −4 1 .
ARS = AR AS = 1 −1 1 0
1 0 −1
−1 −2 1
2 1 −1
Check that ARS v1 = w1 , ARS v2 = w2 , ARS v3 = w3 , by multiplying the
matrices:
1 0
0
1
1
ARS v1 = 0 −4 1 0 = 1 = w1 ,
2 1 −1
1
1
1
1
1 0
0
ARS v2 = 0 −4 1 1 = −1 = w2 ,
0
3
2 1 −1
0
0
1 0
0
ARS v3 0 −4 1 0 = 1 = w3
−1
2 1 −1
1
Problem 7.3 For each of the linear transformations a basis of the null space,
N (T ), is obtained as a basis of the null space of the associated matrix AT , and a
basis of R(T ) is a basis of the range, or column space of the matrix AT .
(a) T : R2 → R3 ,
( )
x + 2y
x
T
= 0 ,
y
0
1
AT = 0
0
2
0
0
CHAPTER 7.
70
This matrix is in reduced row echelon form.
(
)
−2
basis of N (T ) :
∈ R2 ,
1
1
basis of R(T ) :
0 ∈ R3 ,
0
nullity(T ) = 1
rank(T ) = 1 .
For the rank-nullity theorem, we have
rank(T ) + nullity(T ) = 1 + 1 = 2 = dim(R2 ).
This transformation is not invertible since the spaces have different dimensions.
(b) T : R3 → R3 ,
x
x+y+z
T y = y + z ,
z
z
1
AT = 0
0
1
1
0
1
1.
1
The reduced row echelon form of the matrix is
1 0 0
AT = 0 1 0 .
0 0 1
Therefore N (T ) = {0}, so nullity(T ) = 0, and R(T ) = R3 , so rank(T ) = 3.
Then
rank(T ) + nullity(T ) = 3 + 0 = 3 = dim(R3 ).
This transformation is invertible and AT −1 = A−1
T . Using either row operations
or the cofactor method,
u−v
1 −1 0
u
0 1 −1 , so T −1 v = v − w .
A−1
T =
w
w
0 0
1
Check that AT −1 AT = I or that T −1 T = I.
(c) T : R3 → R3 ,
x
1 1 0
x
T y =
0 1 1
y = AT x
z
−1 0 1
z
71
Reduce the matrix to reduced row echelon form,
1 1 0
1 0
0 1 1 → ··· → 0 1
−1 0 1
0 0
basis of N (T ) :
basis of R(T ) :
−1
1 .
0
1
−1 ⊂ R3 ,
nullity(T ) = 1
1
.
1
1
0 1 ⊂ R3 ,
rank(T ) = 2
−1
0
The rank-nullity theorem is:
rank(T ) + nullity(T ) = 2 + 1 = 3 = dim(R3 ).
This transformation is not invertible since dim(N (T )) ̸= 0. (Equivalently, AT is
not invertible.)
Problem 7.4 Let T : R3 → R4 be the linear transformation defined by
T (e1 ) = v1 ,
T (e2 ) = v2 ,
The matrix AT , such that T (x) = AT x, is
1
2
AT =
0
−1
T (e3 ) = v3 .
1 4
1 5
.
1 3
1 2
The range of T is the column space of the matrix AT , so to answer this question,
you need to know how many vectors are contained in a basis of
R(T ) = CS(AT ) = Lin{v1 , v2 , v3 }. For this, reduce the matrix AT to row
echelon form. (Here the reduced row echelon form is also shown.)
1 0 1
1 1 4
1 1 4
0 1 3
0 1 3
2 1 5
A=
0 1 3 → ··· → 0 0 0 → ··· → 0 0 0.
0 0 0
0 0 0
−1 1 2
CHAPTER 7.
72
Since there are only two leading ones, the rank of the matrix is two and
dim(R(T )) = 2.
To determine if w ∈ R(T ), you need to establish if it is a linear combination of
the basis vectors of R(T ). From the row echelon form, you can deduce that a
basis of R(T ) is {v1 , v2 }. If you set w = sv1 + tv2 , and try to solve for s and t,
you will find that equating components gives four equations in two unknowns
and the system is inconsistent. Therefore w is not a linear combination of the
basis vectors and is therefore not in R(T ).
Next, you are asked to state the rank-nullity theorem for linear transformations
and use it to determine the dimension of the null space of T , N (T ).
The rank-nullity theorem is
dim(R(T )) + dim(N (T )) = n,
where T maps from a vector space of dimension n. The important thing to
remember is to state (or indicate) that n is the dimension of the domain. Here
n = 3 and dim(R(T )) = 2, so dim(N (T )) = 1.
For a basis of N (T ), look at the reduced row echelon form of AT (shown above)
and write down a basis of the null space. The solutions of AT x = 0 are
−1
x1
−t
x = x2 = −3t = t −3 , t ∈ R,
1
x3
t
so a basis of N (T ) consists of the vector v = (−1, −3, 1)T . (Note that, as
T : R3 → R4 , the null space of T is a subspace of R3 .)
Problem 7.5 This question requires the use of the rank-nullity theorem:
dim(R(T )) + dim(N (T )) = n
where n is the dimension of the domain space.
For T1 , both the null space and the range have dimension one,
dim(R(T )) + dim(N (T )) = 1 + 1 = 2,
73
but the given domain is R3 with dimension 3. Therefore no such linear
transformation can exist.
T2 is trickier. Since the domain is R2 , we have n = 2 in the rank-nullity theorem.
The dimension of the null space is 0. What is the dimension of the range? The
three given vectors span R(T2 ), but are they linearly independent? You can check
this using the matrix whose columns are these three vectors, or you may notice
that the third vector is the sum of the first two. So the linear span of these three
vectors is a two-dimensional subspace of R3 , and the rank-nullity theorem does
not rule out the possibility that there is such a linear transformation. So to show
that it does exist, find a suitable matrix. What size must the matrix be?
AT2 must be 3 × 2, so one choice is
1 1
AT2 = 0 1 .
1 1
For this matrix, R(AT2 ) is the required subspace of R3 and N (AT2 ) = {0}.
For T3 , both the null space and the range are lines (in different Euclidean spaces),
but each has dimension one. Since the domain is R2 , such a linear transformation
would satisfy the rank-nullity theorem. To find a matrix representing T , find a
basis for each of the subspaces. We can express them as
{ ( )
}
−4
1
N (T3 ) = t
t∈R
and R(T3 ) = t 5 t ∈ R .
2
1
Since T3 : R2 → R3 , the matrix AT3 must be 3 × 2. If the columns are c1 and c2 ,
then the columns must both be in the range, and the linear dependence relation
must be given by the basis of the null space; that is, c1 + 2c2 = 0, or c1 = −2c2 .
So one possible choice of the matrix is
8
−4
AT3 = −10 5 .
−1
1
Problem 7.6 The rank-nullity theorem for linear transformations is
dim(N (T )) + dim(R(T )) = n,
CHAPTER 7.
74
where n is the dimension of the domain of T . Both V and W have dimension
two, so we must have n = 2 + 2 = 4. Therefore the domain of the linear
transformation must be R4 , and not R3 . Since the null space must be a subspace
of the domain, the linear transformation T cannot therefore exist.
A matrix of a linear transformation satisfying the conditions of S : R4 → R3 ,
must be 3 × 4. Let c1 , c2 , c3 , c4 denote the columns of AS . We can take the first
two columns to be the basis vectors of R(T ) = V ,
1
1
AS = 0 −1 c3 c4 .
−1 0
Since N (S) = W , we use the basis vectors of W to obtain the linear dependence
relations of the column vectors. Then
1
−4
1
1c1 + 3c2 + c3 = 0 =⇒ c3 = − 0 − 3 −1 = 3
1
0
−1
−3
1
1
−2c1 + 5c2 + c4 = 0 =⇒ c4 = 2 0 − 5 −1 = 5 .
−2
0
−1
Therefore,
1
1 −4 −3
AS = 0 −1 3
5
−1 0
1 −2
is one possible solution. Putting this matrix into reduced row echelon form will
confirm that N (S) = W and R(S) = V .
Problem 7.7 Put the matrix AT into reduced row echelon form:
1 −3
1 −3
AT = −2 6 −→ 0 0
−1 3
0 0
so
{( )}
3
N (T ) = Lin
1
and
1
R(T ) = Lin −2 .
−1
75
Reducing AS ,
1
AS = −1
0
0 1
1 0
1 0 → ··· → 0 1
2 1
0 0
1
1
1
so that N (S) = {0} and R(S) = R3 .
Only the linear transformation ST is defined, ST : R2 → R3 .
Since N (S) = {0}, N (ST ) = N (T ). To see this, simply note that
x ∈ N (ST )
⇐⇒
⇐⇒
⇐⇒
⇐⇒
S(T (x)) = 0
T (x) ∈ N (S) = {0}
T (x) = 0
x ∈ N (T ).
So N (ST ) is the one-dimensional subspace of R2 with basis v = (3, 1)T . By the
rank-nullity theorem,
dim(R(ST )) + dim(N (ST )) = n = dim(R2 ) = 2,
so dim(R(ST )) = 1.
Problem 7.8 (a) To find a basis of the null space and the range of T , you reduce
the matrix A. But reading ahead, you can reduce the augmented matrix (A|d).
We have
5 9 −1 15 4
1 0 −2 0 −1
A = −3 1 7 −4 4 → · · · → 0 1 1 0 1 .
1 2 0
3 1
0 0 0 1 0
Therefore Ax = d is consistent, so d ∈ R(T ). Reading the solutions of Ax = d
from the reduced row echelon form, a general solution is
2
−1
1
+ t −1 = p + tv, t ∈ R,
x=
1
0
0
0
CHAPTER 7.
76
and we can conclude from this that a basis of the null space is {v}. Looking at
the reduced row echelon form of A (the first four columns), we see that every
row has a leading one, so R(T ) is a 3-dimensional subspace of R3 , so that
R(T ) = R3 .
Alternatively, you can reduce the matrix A to obtain the basis of the null space,
and determine that R(T ) = R3 . and therefore d ∈ R(T ). Then if you notice that
d = c2 − c1 where ci denote the columns of A, you can deduce that a particular
solution is p = (−1, 1, 0, 0)T and obtain the same result for x.
(b) Now suppose S : R3 → R2 , is such that the range of S is R2 and the null
space of S is the subspace:
N (S) = {x | x = td, t ∈ R},
where d = (4, 4, 1)T is the vector given above. The linear transformation ST
T
S
maps R4 −→R3 −→R2 . Since the range of T is all of R3 , the range of ST is
equal to the range of S, which is R2 . The rank-nullity theorem then tells us that
dim(R(ST )) + dim(N (ST )) = n,
where n is the dimension of the domain. Since this is R4 , n = 4, so
2 + dim(N (ST )) = 4, and dim(N (ST )) = 2.
To find a basis of N (ST ), we need to find two linearly independent vectors in R4
which are mapped to 0. One of them is the basis vector of the null space of T .
The other is the vector p. Why? Since T (p) = d and S(d) = 0, ST (p) = 0.
These two vectors are linearly independent (since they are not scalar multiples of
one another). Therefore the set {v, p} is a basis of N (ST ).
You could also find a matrix representing S using the information given about
the range and null space. For example
(
)
1 0 −4
AS =
,
0 1 −4
defines such a linear transformation S. Then you can calculate AST = AS AT to
obtain the same results.
77
Problem 7.9 The determinant of P is
1 0
|P | = 1 1
3 −2
1
2 = 3.
4
Since |P | ̸= 0, the columns of P form a basis B = {c1 , c2 , c3 } of R3 . Why?
Since |P | ̸= 0, P −1 exists and P x = b has a unique solution for all b ∈ Rn
(namely, x = P −1 b). This says that any b ∈ Rn is a unique linear combination
of the column vectors of P ; that is, B spans R3 and B is linearly independent.
If w = (1, 1, 0)T in standard coordinates, then to find [w]B you either need to
solve a system of equations using Gaussian elimination or use the matrix P −1 ,
since x = P [x]B . The matrix P −1 can be found using either the adjoint matrix
(cofactors method) or by row reduction of (P |I). We have
8 −2 −1
1
1 −1 ,
P −1 = 2
3
−5 2
1
which you should immediately check by calculating that P P −1 = I.
Then
1
2
8 −2 −1
1
[w]B = P −1 w = 2
1 −1 1 = 1 .
3
0
−1 B
−5 2
1
Check this by confirming that
1
0
1
1
1
− 2 = 1.
w=2 1 +
4
0
3
−2
If
6
[v]B = −1 ,
−2 B
then
1
0
1
4
v=6 1 −
1
−2 2 =
1 .
3
−2
4
12
So in standard coordinates, v = (4, 1, 12)T .
CHAPTER 7.
78
b are bases can be checked in one of the usual ways.
Problem 7.10 That B and B
The required transition matrices to standard coordinates are
1 1 0
1 1
0
P = 0 1 0 , Q = 1 −1 1 .
1 3 1
1 0 −1
b coordinates to B coordinates is P −1 Q, where P is
The transition matrix from B
the transition matrix from B coordinates to standard, and Q is the transition
b coordinates to standard. So you first need to write down P and Q
matrix from B
and calculate P −1 . (The calculations are not shown here.) Then
1 −1 0
1 1
0
0
2 −1
P −1 Q = 0
1 0 1 −1 1 = 1 −1 1
−1 −2 1
1 0 −1
−2 1 −3
Computing [x]B using this matrix,
−1
2
0
2 −1
[x]B = 1 −1 1 1 = 4
−12 B
3 Bb
−2 1 −3
You could check your solution by finding x in standard coordinates from both
bases.
Problem 7.11 Rotation of the axes through an angle of θ = π/6 clockwise (see
Example 7.16) is accomplished by the linear transformation whose matrix is
(
) ( √3
1 )
cos(π/6) sin(π/6)
2
2
√
=
= P.
− sin(π/6) cos(π/6)
− 12 23
The new basis vectors are
( √3 )
v1 =
2
− 12
( 1 )
,
v2 =
√2
3
2
.
The transition matrix from B coordinates to standard coordinates is the matrix P .
To change coordinates, we write x = P [x]B . We have
√
3
1
√
( ) ( 3
)
X
+
Y
x
=
1 )(
x
X
2
2
2
√2
=
=⇒
√
y
Y
− 12 23
y = −1X + 3Y .
2
2
79
√
Substituting for x and y in the equation for C, 6 = 3x2 + 2 3xy + 5y 2 ,
(√
6=3
3
1
X+ Y
2
2
)2
√
+2 3
(√
3
1
X+ Y
2
2
)(
√ ) (
√ )2
1
3
1
3
− X+
Y +5 − X +
Y
.
2
2
2
2
Multiplying out and collecting terms, this reduces to X 2 + 3Y 2 = 3, which is
the equation of the curve C in the new B coordinates. This is the equation of an
ellipse, centered at zero, with its major axis on the new X axis and its minor axis
on the new Y axis. The positions of the new axes are found by rotating the old x
and y axes through an angle of π6 radians clockwise. The ellipse intersects the X
√
axis at (± 3, 0) and the Y axis at (0, ±1), as shown in the following figure.
y
√
3
Y
1
x
−1
√
− 3
X
Problem 7.12 (a) M is a basis of R2 since the transition matrix P has non-zero
determinant:
(
)
2 −1
P =
,
|P | = 3 ̸= 0.
1 1
P is the transition matrix from M coordinates to standard coordinates.
Since v = P [v]M , we have [v]M = P −1 v:
(
)(
) [
]
1
1 1
−1
1/3
[v]M =
=
.
2
5/3 M
3 −1 2
CHAPTER 7.
80
(b) The matrix of the linear transformation T with respect to the standard basis is
(
)
7 −2
A=
.
−1 8
The matrix of T in M coordinates is
)(
)(
) (
)
(
1
7 −2
2 −1
6 0
1 1
−1
=
.
D = P AP =
−1 8
1 1
0 9
3 −1 2
Since
(
6 0
0 9
)[ ]
[ ]
1
1
=6
0 M
0 M
(
and
6 0
0 9
)[ ]
[ ]
0
0
=9
,
1 M
1 M
the linear transformation T is a stretch in the direction v1 = (2, 1)T by a factor of
6, and a stretch in the direction v2 = (−1, 1)T by a factor of 9.
(c) To find the image of [v]M using D,
[ ]
(
)[
]
2
6 0
1/3
.
=
[T (v)]M = D[v]M =
15 M
0 9
5/3 M
To check, compute T (v) using standard coordinates and show that it is equal to
[T (v)]M , expressed in standard coordinates:
(
)(
) (
)
7 −2
−1
−11
T (v) =
=
,
−1 8
2
17
and
( )
(
) (
)
2
−1
−11
2
+ 15
=
.
1
1
17
Chapter 8
Problem 8.1 Multiply the matrix A with each of the vectors.
0
1
1 1 0
−1 = 6 ,
Ax = 1 4 3
0
3
0 3 1
which is not a scalar multiple of x, so x is not an eigenvector.
1
6
1
1 1 0
5 = 30 = 6 5 ,
Ay = 1 4 3
3
18
3
0 3 1
so y is an eigenvector (with corresponding eigenvalue λ = 6).
1 1 0
5
5
Az = 1 4 3 0 = 8 ,
0 3 1
1
1
so z is not an eigenvector.
Problem 8.2 The characteristic equation is
|A − λI| =
1−λ
3
4
= λ2 − 3λ − 10 = (λ − 5)(λ + 2) = 0
2−λ
so the eigenvalues are λ = 5 and λ = −2.
Find corresponding eigenvectors by solving (A − λI)v = 0.
81
CHAPTER 8.
82
For λ1 = 5,
(
A − 5I =
−4 4
3 −3
so the corresponding eigenvectors are
( )
1
v=t
,
1
)
(
−→
1 −1
0 0
)
t ∈ R, t ̸= 0.
For the eigenvectors for λ2 = −2, we have
)
(
(
)
3 4
1 34
A + 2I =
→
=⇒
3 4
0 0
(
v=t
−4
3
)
,
t ∈ R, t ̸= 0.
Then taking one eigenvector for each eigenvalue, we can set
(
)
1 −4
P =
.
1 3
Then P −1 AP is a diagonal matrix D, with the corresponding eigenvalues on the
main diagonal:
(
)
5 0
D=
.
0 −2
To show P −1 AP = D, first multiply AP to check the eigenvectors are correct
(the columns of AP should be λi vi ).
(
)(
)(
)
(
)(
)
1
1
3 4
1 4
1 −4
3 4
5 8
−1
=
P AP =
3 2
1 3
5 −6
7 −1 1
7 −1 1
(
) (
)
1 35
0
5 0
=
=
= D.
0 −2
7 0 −14
Problem 8.3 As the method is the same as in the previous problem, we will not
show details of the calculations. The eigenvalues are λ1 = 6 and λ2 = 9 with
corresponding eigenvectors v1 = (2, 1)T and v2 = (−1, 1)T . If
(
)
(
)
2 −1
6 0
P =
and D =
, then P −1 AP = D.
1 1
0 9
The linear transformation T is a stretch by a factor of 6 in the direction of v1 (so
any point on the line x = tv1 , t ∈ R maps to another point on the same line,
T (x) = A(tv1 ) = 6(tv1 ) = 6x), and T is a stretch by a factor of 9 in the
direction of v2 . (Compare this with Problem 7.12.)
83
Problem 8.4 The characteristic equation is |B − λI| = 0. Calculating the
determinant, we have
3−λ
−1
|B − λI| =
5
−3 − λ
1
−1
2
5
2−λ
= (3 − λ)(−6 + λ + λ2 + 5) + 1(10 − 5λ − 5) + 2(−5 + 3 + λ)
= −λ3 + 2λ2 + 4λ − 3 − 5λ + 5 + 2λ − 4
= −λ3 + 2λ2 + λ − 2 = 0.
You need to factorise the polynomial λ3 − 2λ2 − λ + 2, and it is usefully given
that the roots are integers. Since the product of the roots is equal to the constant
term, −2, try λ = 1. Substitute λ = 1 into the characteristic equation. As the
equation is satisfied, it is a root and (λ − 1) is a factor, so the polynomial
factorises as λ3 − 2λ2 − λ + 2 = (λ − 1)(λ2 + aλ − 2), where the constant a is
obtained by comparing terms. We find
λ3 − 2λ2 − λ + 2 = (λ − 1)(λ2 − λ − 2) = (λ − 1)(λ − 2)(λ + 1) = 0.
The eigenvalues are 1, 2, −1. Next find an eigenvector for each eigenvalue. For
λ1 = 1 :
1 −1 1
1 −1 1
2 −1 2
1 0 1
B−I = 5 −4 5 → 5 −4 5 → 0 1 0 → 0 1 0
0 0 0
0 1 0
2 −1 2
1 −1 1
−1
=⇒
v1 = 0
1
For λ2 = 2 :
1 −1 2
1 −1 2
1 −1 0
B − 2I = 5 −5 5 → 0 0 −5 → 0 0 1
1 −1 0
0 0 −2
0 0 0
1
=⇒ v2 = 1
0
CHAPTER 8.
84
For λ3 = −1 :
4 −1
B+I = 5 −2
1 −1
2
1 −1
5 → 4 −1
3
5 −2
=⇒
3
1 −1
3
1 0 − 31
2 → 0 3 −10 → 0 1 − 10
3
0 0
0
5
0 3 −10
1
v3 = 10
3
The set of vectors {v1 , v2 , v3 } is a basis of R3 consisting of eigenvectors of B.
Write the three eigenvectors as the columns of the matrix P (in any order). For
example, if
1 0 0
−1 1 1
D = 0 2 0 , then P −1 AP = D.
P = 0 1 10 ,
0 0 −1
1 0 3
Before computing P −1 , check that BP = P D = ( 1v1 2v2
3 −1 2
−1 1 1
−1 2
BP = 5 −3 5 0 1 10 = 0 2
1 −1 2
1 0 3
1 0
1
−1 2 −1
3 −3 9
1
−1
P BP =
0 2 −10 = 0
10 −4 10
6
0
1 0 −3
−1 1 −1
(−1)v3 ):
−1
−10 .
−3
0 0
2 0 = D.
0 −1
Problem 8.5 The eigenvalues of the matrix A are easily obtained by expanding
the determinant |A − λI| by the middle column. The eigenvalues are 1, −1 and
7. Using the same method as in the previous problem, find an eigenvector for
each eigenvalue and write these as the columns of a matrix P . One possible
choice is:
−2 0 4
1 0 0
P = 1 1 1,
D = 0 −1 0 .
2 0 2
0 0 7
Then P −1 AP = D. You should check the eigenvectors by showing that
AP = P D.
85
Problem 8.6 The characteristic equation is
|C − λI| =
5−λ
0
a
−1 − λ
2
0
4
b
= 0.
3−λ
Expanding the determinant by the second column,
|C−λI| = (−1−λ)(λ2 −8λ+15−8) = (−1−λ)(λ2 −8λ+7) = (−1−λ)(λ−1)(λ−7),
so this matrix has three distinct eigenvalues, λ = −1, 1, 7 for any values of
a, b ∈ R. Therefore it can be diagonalised.
Problem 8.7
|A − λI| =
1−λ
0
1
1
1−λ
0
1
−1 = −λ(λ − 2)2 = 0.
2−λ
The eigenvalues are λ = 0 and λ = 2 (multiplicity two).
There will be one eigenvector for λ = 0. Determine if two linearly independent
eigenvectors can be obtained for λ = 2.
−1 1
1
1 0 0
(A − 2I) = 0 −1 −1 −→ · · · −→ 0 1 1 .
1
0
0
0 0 0
The dimension of the null space of this matrix is dim(N (A − 2I)) = 1, so there
is only one linearly independent eigenvector for λ = 2, and therefore the matrix
A cannot be diagonalised.
|B − λI| =
−2 − λ 1
−1
−λ
2
1
−2
= −λ3 + 8 = 0,
1
2−λ
so λ = 2 is the only real root. The other two roots are complex, so the matrix
cannot be diagonalised over the real numbers.
86
CHAPTER 8.
Problem 8.8 Using the definition of eigenvector,
−5 8 32
2
6
2
Av = 2 1 −8 −2 = −6 = 3 −2 ,
−2 2 11
1
3
1
so v is an eigenvector of A corresponding to the eigenvalue λ = 3.
To find all eigenvectors for this eigenvalue, solve (A − 3I)x = 0. Putting the
matrix (A − 3I) into reduced row echelon form,
−8 8
32
1 −1 −4
A − 3I = 2 −2 −8 → 0 0
0 .
−2 2
8
0 0
0
This matrix has rank 1, so there are two linearly independent eigenvectors
corresponding to λ = 3. The solutions are
4
1
s + 4t
= s 1 + t 0 = sv1 + tv2 , s, t ∈ R.
x=
s
1
0
t
Therefore the eigenspace is the plane Lin{v1 , v2 } ⊂ R3 .
Note that the vector v ∈ Lin{v1 , v2 } since it is an eigenvector for this
eigenvalue: in fact, v = v2 − 2v1 .
To find the remaining eigenvalue and a corresponding eigenvector, you can find
the characteristic equation. Expanding the determinant of A − λI by the first
row, for example, you will find a common factor of (3 − λ) in all three terms, so
that you do not need to multiply out the cubic equation. Alternatively, you can
find |A| = 9, which (by Theorem 8.11) is the product of the eigenvalues. So the
remaining eigenvalue is λ = 1. A corresponding eigenvector is found by solving
(A − I)x = 0, with solutions x = t(4, −1, 1)T , t ∈ R. If we take
1 4 4
3 0 0
P = 1 0 −1 , D = 0 3 0 ,
0 1 1
0 0 1
then P −1 AP = D and we have diagonalised A.
87
Problem 8.9 (a) To verify that v1 is an eigenvector, multiply Av1 :
4 3 −7
1
3
1 2 1 2 = 6,
2 2 −3
1
3
which shows that v1 is an eigenvector corresponding to the eigenvalue λ1 = 3.
To diagonalise A we need to find the characteristic equation.
4−λ
1
2
3
−7
= −λ3 + 3λ2 + λ − 3 = 0.
2−λ
1
2
−3 − λ
We know that λ = 3 is a root, so we find that
λ3 − 3λ2 − λ + 3 = (λ − 3)(λ2 − 1). The eigenvalues are 3, 1, −1.
The eigenvectors for λ2 = 1 and λ3 = −1 are found by solving (A − I)x = 0
and (A + I)x = 0 respectively. So for λ2 = 1 we can take the eigenvector
v2 = (−1, 1, 0)T , and for λ3 = −1 we have v3 = (2, −1, 1)T . Then if
3 0 0
1 −1 2
P = 2 1 −1 , and D = 0 1 0 , then P −1 AP = D.
0 0 −1
1 0
1
Check that AP = P D:
4 3 −7
1 −1 2
3 −1 −2
(
)
1 2 1 2 1 −1 = 6 1
1
= 3v1 1v2 (−1)v3 .
2 2 −3
1 0
1
3 0 −1
(b) The determinant of A is the product of the eigenvalues, so |A| = −3 ̸= 0.
Therefore, A is invertible.
Solving P −1 AP = D for A, we have A = P DP −1 , and therefore
A−1 = (P DP −1 )−1 = P D−1 P −1
1
So
P −1 (A−1 )P = D−1 ,
3
where D−1 = 0
0
0 0
1 0 .
0 −1
CHAPTER 8.
88
So P is also the matrix of eigenvectors for A−1 , and D−1 is the corresponding
diagonal matrix.
(c) The missing entries are
s12 = C21 = (−1)3
3 −7
= −5 and
2 −3
s31 = C13 = (−1)4
1
2
2
= −2.
2
Then checking the eigenvectors by multiplying A−1 P ,
−8 −5 17
1 −1 2
−1 3
6
1
1
− 5
2 −11 2 1 −1 = − −2 −3 −3 = P D−1 .
3
3
−2 −2
5
1 0
1
−1 0
3
Problem 8.10 Each of the matrices A and B has the same set of eigenvalues:
λ = −1 and λ = 5 of multiplicity 2. (These are easily found from the
characteristic equations.) To see if either matrix can be diagonalised, you need to
check whether you can find two linearly independent eigenvectors for λ = 5.
For the matrix A,
1 −1
−3 3 0
A − 5I = 3 −3 0 → 1 1
0 0
1
1 0
0
0,
0
so this matrix does not have two linearly independent eigenvectors for λ = 5 and
cannot be diagonalised.
For the matrix B,
1 −1 0
s
−3 3 0
B − 5I = 3 −3 0 → 0 0 0 =⇒ x = s ,
t
1 −1 0
0 0 0
s, t ∈ R.
Therefore the eigenspace for λ = 5 is Lin{v1 , v2 } with v1 = (1, 1, 0)T and
v2 = (0, 0, 1)T . An eigenvector corresponding to λ3 = −1 is v3 = (−3, 3, 1)T .
Therefore the matrix B can be diagonalised: if
1 0 −3
5 0 0
P = 1 0 3 , D = 0 5 0 ,
0 1 1
0 0 −1
then P −1 AP = D.
89
Problem 8.11 If such a linear transformation existed, then there would be a
3 × 3 matrix A such that T (x) = Ax. The description of T indicates that
Av1 = 2v1 ,
Av2 = v2 ,
Av3 = 0v3 = 0.
Therefore the matrix A would have the distinct eigenvalues 2, 1 and 0, and their
corresponding eigenvectors would therefore be linearly independent. But the
vectors v1 , v2 , v3 are linearly dependent since v3 = v1 + v2 . Therefore no such
linear transformation can exist.
Problem 8.12 If A and B are diagonalisable n × n matrices with the same
eigenvalues, then there are invertible matrices P and Q, and a diagonal matrix D
such that P −1 AP = D and Q−1 BQ = D. Then
P −1 AP = D = Q−1 BQ
This can be written as
=⇒
QP −1 AP Q−1 = B.
(P Q−1 )−1 A(P Q−1 ) = B.
Therefore A and B are similar matrices: we have B = S −1 AS with S = P Q−1
Problem 8.13 Calculating the characteristic polynomial, the eigenvalues of A
are λ = 1, 3. Finding an eigenvector for each eigenvalue, the matrix is
diagonalised by taking
(
)
(
)
−1 −2
1 0
P =
, D=
.
1
1
0 3
Then P −1 AP = D.
The matrix B has the same eigenvalues, and is diagonalised by taking, for
example,
(
)
(
)
4 3
1 0
Q=
, D=
.
1 1
0 3
Then Q−1 BQ = D.
Then, as we saw in the previous problem, taking S = P Q−1 , we have
S −1 AS = B. Check by multiplying the matrices:
(
)(
) (
)
−1 −2
1 −3
1 −5
−1
S = PQ =
=
.
1
1
−1 4
0 1
CHAPTER 8.
90
Then
(
S
−1
AS =
1 5
0 1
)(
5
4
−2 −1
)(
1 −5
0 1
)
(
=
−5 24
−2 9
)
= B.
Problem 8.14 Matrices A and AT have the same characteristic polynomial,
because by properties of transpose and determinant,
|A − λI| = |(A − λI)T | = |AT − λI|.
Since the eigenvalues are the roots of the characteristic polynomial, the matrices
have the same eigenvalues.
Chapter 9
Problem 9.1 The matrix A was diagonalised in Problem 8.2. We found
P −1 AP = D, where
(
)
(
)
(
)
1
3 4
1 −4
5 0
−1
.
P =
, D=
, P =
1 3
0 −2
7 −1 1
Solving for A, we have A = P DP −1 , and therefore A5 = P D5 P −1 .
(
)(
)5 (
)
1
1 −4
5 0
3 4
5
A =
1 3
0 −2
7 −1 1
(
)(
) (
) (
).
1
1 −4
3125
0
3 4
1321 1804
=
=
1 3
0
−32 7 −1 1
1353 1772
Problem 9.2 In matrix form, the system of difference equations is xt+1 = Axt
where A is the matrix in the previous problem,
(
)
1 4
A=
.
3 2
Using the same diagonalisation, the solution to xt+1 = Axt is given by
xt = P Dt P −1 x0 , so that
( ) (
)( t
) (
)( )
1
1 −4
xt
5
0
3 4
1
=
.
t
yt
1 3
0 (−2) 7 −1 1
0
The sequences are
4
3
xt = (5t ) + (−2)t
7
7
3 t
3
yt = (5 ) − (−2)t .
7
7
91
CHAPTER 9.
92
The values of x5 and y5 can be found from these expressions using t = 5, or
using the result of the previous question. Since xt = At x0 ,
( ) (
)( )
1321 1804
1
x5
=
,
1353 1772
0
y5
so x5 = 1321 and y5 = 1353. (You can check your work by obtaining this result
both ways.)
Problem 9.3 This system of difference equations is xt+1 = Axt where A is the
matrix you diagonalised in Problem 8.9, and xt = (xt , yt , zt )T . We have
1 −1 2
3 0 0
P = 2 1 −1 , D = 0 1 0 , and P −1 AP = D.
1 0
1
0 0 −1
Therefore, the solution is given by xt = P Dt P −1 x0 where x0 = (4, 5, 1)T . Find
P −1 using cofactors or row operations. Then
4
1
1 −1
8
1
1
P −1 x0 = −3 −1 5 5 = −12 ,
2
2
1
−1 −1 3
−6
and
t
3 0
0
4
xt
1 −1 2
0 −6 ,
xt = yt = 2 1 −1 0 1t
0 0 (−1)t
−3
zt
1 0
1
4(3)t
xt
1 −1 2
yt = 2 1 −1 −6 .
zt
1 0
1
−3(−1)t
The sequences are
xt = 4(3t ) + 6 − 6(−1)t
yt = 8(3t ) − 6 + 3(−1)t
zt = 4(3t ) − 3(−1)t .
Checking the result, first note that it satisfies x0 = 4, y0 = 5 and z0 = 1.
Substituting into the expressions,
x1 = 12 + 6 + 6 = 24,
y1 = 24 − 6 − 3 = 15,
z1 = 12 + 3 = 15,
93
which agrees with x1 = Ax0 :
x1
4 3 −7
4
24
y1 = 1 2 1 5 = 15 .
z1
2 2 −3
1
15
Problem 9.4 To solve the system of difference equations, xk = Axk−1 ,
diagonalise the matrix A. Then the solutions are xk = P Dk P −1 x0 . The
characteristic equation factorises easily, since we know that λ = 1 is a solution:
|A − λI| =
0.7 − λ
0.3
0.6
= λ2 − 1.1λ + 0.10 = (λ − 1)(λ − 0.1) = 0.
0.4 − λ
Hence the eigenvalues are 1 and 0.1.
For corresponding eigenvectors,
(
)
(
)
( )
−0.3 0.6
1 −2
2
λ=1:
A−I =
−→
=⇒ v1 =
0.3 −0.6
0 0
1
(
A − 0.1I =
λ = 0.1 :
0.6
0.3
0.6
0.3
)
(
If
P =
then
P
and
xk =
(
2 −1
1 1
−1
1
x0 =
3
)( k
1
0
(
1 1
−1 2
0
(0.1)k
)
1
3
−→
2 −1
1 1
)(
(
As k → ∞,
(
1
0.2
(
xk
→
)
(
=⇒ v2 =
−1
1
)
)
,
0.6
0.4
)
1 1
0 0
)
1
=
3
1
=
3
2/3
1/3
)
.
(
1
0.2
)
,
(
( )
)
1
2
−1
k
+ (0.1)
.
1
1
15
CHAPTER 9.
94
Problem 9.5 A is the transition matrix of a Markov process since
(1) all entries are non-negative, (2) the entries of each column sum to 1.
Therefore λ = 1 is an eigenvalue and all other eigenvalues satisfy |λi | ≤ 1.
Solving the characteristic equation, |A − λI| = 0, the eigenvalues are
λ1 = 1, λ2 = 12 , λ3 = − 12 . Solving (A − λI)v = 0 for each of these, you should
find that corresponding eigenvectors are
1
0
−2
v1 = 1 , v2 = −1 , v3 = 1 .
1
1
1
If
1 0 −2
P = 1 −1 1
1 1
1
and
1
D= 0
0
0
1
2
0
0
0 ,
− 12
then P −1 AP = D. (You should check that AP = P D.)
The solution to xt = Axt−1 is given by xt = P Dt P −1 x0 , but since you are only
asked for the long term distribution, you only need to find the eigenvector for
λ = 1 which is also a distribution vector (meaning its components sum to one).
Therefore, as t → ∞, we have
1
1
xt →
1
3
1
and the long term population distribution is 2000 members in each of the three
states. (Notice that you did not need to know x0 to answer this question.)
Problem 9.6 (a) This is not a Markov process since not all entries of the matrix
A are non-negative. Also, the columns do not add up to 1.
In each year, the population of osprey is given by 60% of the previous year’s
osprey population plus 20% of the previous year’s trout population. In each year,
the population of trout decreases by 25% of the previous year’s osprey
population and increases by 120% of the previous year’s trout population.
95
Since xt = Axt−1 , we have xt = At x0 , where x0 = (20, 100)T . To solve this
system, you need to diagonalise A. The characteristic equation is
|A − λI| =
0.6 − λ
−0.25
0.2
= λ2 − 1.8λ + 0.77 = (λ − 1.1)(λ − 0.7) = 0.
1.2 − λ
Hence the eigenvalues are λ = 1.1, 0.7. For corresponding eigenvectors,
(
)
(
)
( )
−0.5 0.2
1 − 25
2
λ = 1.1 :
A − 1.1I =
−→
=⇒ v1 =
−0.25 0.1
5
0 0
(
λ = 0.7 :
A − 0.7I =
−0.1 0.2
−0.25 0.5
(
Let
P =
)
(
−→
2
5
2
1
1 −2
0 0
)
( )
2
=⇒ v2 =
1
)
.
Set xt = P zt , so that
z0 = P
−1
1
x0 = −
8
(
1 −2
−5 2
)(
20
100
)
1
=
8
(
180
−100
)
.
Then the solution is xt = At x0 = P Dt P −1 x0 :
(
)(
) (
)
( )
( )
1
180
100
2 2
(1.1)t
0
180
2
2
t
t
xt =
=
(1.1)
−
(0.7)
.
t
5 1
0
(0.7) 8 −100
5
1
8
8
As t → ∞, (1.1)t → ∞ and (0.7)t → 0. Therefore xt → ∞. Both populations
increase without bound in the ratio of 2:5 (osprey to trout).
(b) The transition matrix of a Markov process has λ = 1 as its largest eigenvalue.
All others satisfy |λ| ≤ 1. If the other eigenvalues satisfy |λ| < 1, then as
t → ∞, xt converges to a finite long-term distribution.
So let’s require that the largest eigenvalue of B is 1, and determine the value of α.
|B − λI| =
0.6 − λ
0.2
= λ2 − 1.8λ + (0.72 + 0.2α) = 0
−α
1.2 − λ
CHAPTER 9.
96
If λ = 1 is an eigenvalue, we deduce from the term −1.8λ that this equation
must factorise into
(λ − 1)(λ − 0.8) = 0
⇒
0.72 + .2α = 0.8
=⇒
α = 0.4.
Therefore, if α = 0.4, the eigenvalues are λ = 1 and λ = 0.8.
To demonstrate this works, redo the question using the matrix
)
(
0.6 0.2
.
B=
−0.4 1.2
Diagonalise B:
|B − λI| =
0.6 − λ
−0.4
0.2
= λ2 − 1.8λ + 0.8 = (λ − 1)(λ − 0.8) = 0
1.2 − λ
Hence the eigenvalues are λ = 1, 0.8 (which satisfy the same conditions as
those of a Markov chain).
For corresponding eigenvectors,
(
)
(
)
( )
−0.4 0.2
1 − 12
1
λ=1
B−I =
−→
=⇒ v1 =
−0.4 0.2
0 0
2
(
B − 0.8I =
λ = 0.8
−0.2 0.2
−0.4 0.4
)
(
−→
1 −1
0 0
)
( )
1
=⇒ v2 =
1
Then, using the initial conditions (this is not a Markov chain so use x0 as given),
(
)
(
)(
) (
)
1 1
1 −1
20
80
−1
P =
,
z0 = P x0 = (−1)
=
2 1
−2 1
100
−60
and
(
xt =
1
2
1
1
)(
(1)t
0
0
(0.8)t
)(
80
−60
)
( )
( )
1
1
t
= 80
− 60(0.8)
2
1
As t → ∞, (0.8)t → 0 . The populations are stable, eventually reaching 80
osprey and 160 trout.
97
Problem 9.7 In matrix form this system is y′ = Ay where y = (y1 (t), y2 (t))T
and A is the same matrix as you diagonalised in Problem 8.2. We have
P −1 AP = D where
(
)
(
)
(
)
1 4
1 −4
5 0
A=
, P =
, D=
.
3 2
1 3
0 −2
Define new functions z(t) = (z1 (t), z2 (t))T by setting y = P z. Then y′ = Ay if
and only if z′ = Dz. The solutions of these equations are
)
)(
)
( ) (
( ) (
αe5t
1 −4
y1
αe5t
z1
′
.
, so that y =
=
=
βe−2t
βe−2t
y2
1 3
z2
The general solution is
y1 (t) = αe5t − 4βe−2t
y2 (t) = αe5t + 3βe−2t
α, β ∈ R.
To find the unique solution satisfying the initial conditions y1 (0) = 1 and
y2 (0) = 0, substitute these values into the general solution.
{
( )
( )
(
)( ) ( 3 )
y1 (0) = α − 4β = 1
1
α
1
3 4
1
−1
7
=⇒
=P
=
=
.
β
0
−1
1
0
− 17
7
y2 (0) = α + 3β = 0
The particular solution is
3
4
y1 (t) = e5t + e−2t
7
7
3 5t 3 −2t
y2 (t) = e − e .
7
7
This solution satisfies the initial conditions. The values of y1′ (0) and y2′ (0) can
be found from the original equations,
{
y1′ (0) = y1 (0) + 4y2 (0) = 1(1) + 4(0) = 1
y2′ (0) = 3y1 (0) + 2y2 (0) = 3(1) + 2(0) = 3
or by differentiating the particular solution,
8
15
7
y1′ (t) = e5t − e−2t
y1′ (0) = = 1
7
7
7
=⇒
15
21
6
y ′ (t) = e5t + e−2t
y ′ (0) =
=3
2
2
7
7
7
Note the similarity with the solution to the corresponding system of difference
equations in Problem 9.2.
CHAPTER 9.
98
Problem 9.8 In matrix form, the system is
( ′) (
)( )
3 2
y1
y1
′
y =
=
= Ay.
2 6
y2′
y2
To solve the system, you first need to diagonalise the matrix A. From the
characteristic equation, you should find that the eigenvalues of A are λ1 = 7 and
λ2 = 2. Then find a corresponding eigenvector for each eigenvalue. For example,
if
(
)
1 −2
P =
, D = diag(7, 2), then P −1 AP = D.
2 1
(Check that the eigenvectors are correct by showing that AP = P D.)
To solve the system, define new functions z(t) = (z1 (t), z2 (t))T by setting
y = P z. Then y′ = Ay if and only if z′ = Dz, and z(0) = P −1 y(0). We have
(
)( ) (
)
1
1 2
5
3
z(0) =
=
,
5
−1
5 −2 1
and so the solution of z′ = Dz is
(
z=
3e7t
−e2t
Then the solution of y′ = Ay is
( ) (
) ( 7t )
y1
1 −2
3e
=
,
y2
2 1
−e2t
)
.
or
y1 (t) = 3e7t + 2e2t
y2 (t) = 6e7t − e2t ,
which you can easily check.
Problem 9.9 The characteristic equation of A is
−1 − λ
3
0
2−λ
|A − λI| =
−3
3
0
= −(λ + 1)(λ − 2)2 .
0
2−λ
The eigenvalues are λ = −1 and λ = 2 (with multiplicity two).
99
Find the eigenvectors for λ = 2 first, to see if the matrix can be diagonalised.
Solve (A − 2I)v = 0, by putting the matrix into reduced row echelon form. We
have
−3 3 0
1 −1 0
(A − 2I) = 0 0 0 −→ 0 0 0 ,
−3 3 0
0 0 0
so there will be two linearly independent eigenvectors for λ = 2. Setting the
non-leading variables y = s and z = t, we find the eigenvectors are
x
s
1
0
y = s = s 1 + t 0 , s, t ∈ R.
z
t
0
1
So two linearly independent eigenvectors for λ = 2 are
1
0
v1 = 1 and v2 = 0 .
0
1
For λ = −1,
0
0
(A + I) =
−3
If
1 1
P = 0 1
1 0
3 0
1
3 0 −→ 0
3 3
0
0
0,
1
0 −1
1 0
0 0
−1 0 0
D = 0 2 0,
0 0 2
=⇒
1
v3 = 0
1
then P −1 AP = D.
The system of differential equations is
y1′ (t) = −y1 (t) + 3y2 (t)
y2′ (t) = 2y2 (t)
y3′ (t) = −3y1 (t) + 3y2 (t) + 2y3 (t) .
To find the general solution of the system of linear differential equations
y′ = Ay, set y = P z, then
y′ = Ay ⇐⇒ z′ = Dz.
CHAPTER 9.
100
Solve
′
z1
−z1
z2′ = 2z2
2z3
z3′
to obtain
−t
z1 = αe
z2 = βe2t
z3 = γe2t .
Then y = P z, so the general solution is given by
−t −t
y1
1 1 0
αe
αe + βe2t
y2 = 0 1 0 βe2t =
.
βe2t
2t
−t
2t
y3
1 0 1
γe
αe + γe
Problem 9.10 Write the vectors as the columns of a matrix P and evaluate the
determinant,
1 1 3
P = 0 1 −2 ,
|P | = 1.
1 2 2
Since |P | =
̸ 0, the vectors are linearly independent and are therefore a basis of
3
R.
(a) Since w = P [w]B , we can find the B coordinates using P −1 . Calculating
P −1 using either row operations or cofactors, we have
6
4 −5
2
−5
[w]B = −2 −1 2 −3 = 1 ,
−1 −1 1
1
2 B
which is easily checked.
(b) The matrix P in part (a) has as its columns the eigenvectors for the required
matrix A, and Av1 = v1 , Av2 = 2v2 and Av3 = 3v3 . Therefore,
P −1 AP = D = diag(1, 2, 3). Then solving for A, we have A = P DP −1 .
Multiplying these matrices, we obtain
1 1 3
1 0 0
6
4 −5
−7 −7 8
A = 0 1 −2 0 2 0 −2 −1 2 = 2
4 −2 .
1 2 2
0 0 3
−1 −1 1
−8 −6 9
101
To verify this, show that AP = P D, or Av1 = v1 , Av2 = 2v2 and Av3 = 3v3 :
−7 −7 8
1 1 3
1 2 9
2
4 −2 0 1 −2 = 0 2 −6 = ( 1v1 2v2 3v3 ) .
−8 −6 9
1 2 2
1 4 6
(c) To find the general solution of y′ = Ay, set y = P z, then
y′ = Ay ⇐⇒ z′ = Dz.
Since D = diag(1, 2, 3), the general solution is given by
t t
y1 (t)
1 1 3
αe
αe + βe2t + 3γe3t
.
y = y2 (t) = 0 1 −2 βe2t =
βe2t − 2γe3t
3t
t
2t
3t
y3 (t)
1 2 2
γe
αe + 2βe + 2γe
For the particular solution, note that (α, β, γ)T = z(0) are just the B coordinates
of the vector w. Therefore the unique solution which satisfies y(0) = w is given
by (α, β, γ)T = (−5, 1, 2)T , which is
−5et + e2t + 6e3t
y1 (t)
.
y2 (t) =
e2t − 4e3t
−5et + 2e2t + 4e3t
y3 (t)
(Notice that you can obtain this solution without actually finding A.)
102
CHAPTER 9.
Chapter 10
Problem 10.1 We have b − a = (1, 2, −1, −1)T . The vectors a and b are not
orthogonal, but the vectors a and b − a are orthogonal since
1 ⟩
⟨ 1
1 2
⟨a, b − a⟩ =
2 , −1 = 0,
1
−1
√
√
and they are the same length: ∥a∥ = 7, ∥b − a∥ = 7. In this sense, the three
vectors form an isosceles right triangle. (You can also calculate the angle
between these two vectors using the definition of√cos θ, and you will find that
θ = π/4.) The length of the third side is ∥b∥ = 14. To verify the generalised
Pythagoras theorem with x = a, y = b − a and x + y = b,
∥x + y∥2 = ∥b∥2 = 14 = 7 + 7 = ∥a∥2 + ∥b − a∥2 = ∥x∥2 + ∥y∥2 .
Problem 10.2 (a) Since A is m × k, the matrix AT is k × m, and the product
AT A of a k × m matrix with an m × k matrix is a k × k matrix. To show it is
symmetric, we have
(AT A)T = AT (AT )T = AT A
by properties of transpose. So the matrix is symmetric.
Let x ∈ Rk . Then xT (AT A)x = (Ax)T (Ax) = ∥Ax∥2 > 0 unless Ax = 0.
Since rank(A) = k, the reduced row echelon form of A will have a leading one
in every column (because the columns of A are linearly independent), so the only
103
CHAPTER 10.
104
solution of Ax = 0 is x = 0. Therefore, if x ̸= 0, Ax ̸= 0 and xT (AT A)x > 0.
(Of course, if x = 0, then xT (AT A)x = 0.)
(b) To show that ⟨x, y⟩ = xT (AT A)y defines an inner product, we need to show
that all three properties of the definition are satisfied:
(i) ⟨x, y⟩ = ⟨y, x⟩ for all x, y ∈ V
(ii) ⟨αx + βy, z⟩ = α⟨x, z⟩ + β⟨y, z⟩ for all x, y, z ∈ V and all α, β ∈ R.
(iii) ⟨x, x⟩ ≥ 0 for all x ∈ V , and ⟨x, x⟩ = 0 if and only if x = 0, the zero
vector of the vector space
For property (i), ⟨x, y⟩ = xT (AT A)y = (xT (AT A)y)T since this is just a 1 × 1
matrix. Then
⟨x, y⟩ = xT (AT A)y = (xT (AT A)y)T = yT (AT A)T (xT )T = yT (AT A)x = ⟨y, x⟩
since AT A is symmetric.
Property (ii) follows from properties of matrix algebra:
⟨αx + βy, z⟩ = (αx + βy)T (AT A)z = (αxT + βyT )(AT A)z
= αxT (AT A)z + βyT (AT A)z = α⟨x, z⟩ + β⟨y, z⟩.
Property (iii) follows immediately from part (a). We showed that
⟨x, x⟩ = xT (AT A)x ≥ 0 for all
x ∈ Rk ,
and
⟨x, x⟩ = 0 ⇐⇒ x = 0.
Problem 10.3 If P is an orthogonal matrix,
∥x∥2 = xT x = (P z)T (P z) = zT P T P z = zT z = ∥z∥2
since P T P = I. Therefore, ∥x∥ = ∥z∥.
Problem 10.4 We have
⟨T (x), T (y)⟩ = ⟨P x, P y⟩ = (P x)T (P y) = xT P T P Y = xT y = ⟨x, y⟩.
105
Problem 10.5 The linear transformation T is a rotation anticlockwise by π/4
radians. The linear transformation S is a reflection in the y-axis. You can show
they are orthogonal matrices by multiplying each matrix by its transpose
(ATT AT = I and ATS AS = I) or by noticing that the two columns of each matrix
are orthogonal unit vectors.
Rotation anticlockwise is given by the matrix
)
(
(
cos θ
cos θ − sin θ
T
, so A =
A=
sin θ cos θ
− sin θ
sin θ
cos θ
)
.
Multiply these matrices to show that AT A = I.
Problem 10.6 Write down a basis of the plane
x
y
5x − y + 2z = 0 .
V =
z
by taking any two linearly independent vectors which satisfy the equation. For
example, let
1
0
v1 = 2 , v2 = 5 .
0
1
Then use the Gram-Schmidt process to obtain an orthonormal basis. Let
0
1
u1 = √
2 ,
5 1
and then set
⟨
⟩
1
1
0
0
1
0
1
1
1
w = 5− 5, √ 2 √ 2 = 5−22 = 1 .
5 1
5 1
0
1
−2
0
0
Check that w ⊥ u1 and that w ∈ V . Then
0
1
1
1
{u1 , u2 } = √ 2 , √ 1
5
6 −2
1
CHAPTER 10.
106
is an orthonormal basis of V . You can extend this to an orthonormal basis of R3
using the normal vector to the plane, v3 = (5, −1, 2)T . Then a basis satisfying
the requirements is:
0
1
5
1
1
1
{u1 , u2 , u3 } = √ 2 , √ 1 , √ −1 .
5
6 −2
30
1
2
Problem 10.7 Write the vectors as the columns of a matrix PS and then evaluate
|PS |:
1 2 1
PS = 0 −1 1 , |PS | = −3 ̸= 0.
1 1 5
Therefore S is a basis of R3 . To find an orthonormal basis of the subspace
Lin{v1 , v2 }, let u1 = √12 v1 . Then set
⟨
⟩
1
2
2
2
1
1
1
3 2
1 1
√
−1 , √
w = −1 −
0
0 = −1 −
0 = −1 .
2
2 1
2 1
1
1
1
1
− 12
b = (1, −2, −1)T . Check that w
b ⊥ u1 . (You can also
A vector parallel to w is w
b ∈ Lin{v1 , v2 } by showing that it is a linear combination of v1 and
check that w
v2 .) Then
1
1
1
1
{u1 , u2 } = √ 0 , √ −2
2
6 −1
1
is an orthonormal basis of Lin{v1 , v2 }. (Another possible choice is {u1 , −u2 }.)
You can extend this to a basis of R3 either by applying the Gram-Schmidt
process again using any vector which is not in Lin{v1 , v2 }, such as v3 , or you
can find the Cartesian equation of the plane Lin{v1 , v2 } and use the normal
vector. For example, we have
1
1 2 x
so that n = 1 .
0 −1 y = x + y − z = 0,
−1
1 1 z
107
Then an orthonormal basis B of R3 satisfying the requirements is
1
1
1
1
1
1
{u1 , u2 , u3 } = √ 0 , √ −2 , √ 1 .
2
6 −1
3 −1
1
To find the coordinates of the vectors v2 and v3 in the basis B, you can either
find their coordinates, which are given by their inner products with the basis
vectors, or write down the transition matrix PB from B coordinates to standard
coordinates, and then use [v]B = PB−1 v = PBT v.
Finding the coordinates, since v2 ∈ Lin{v1 , v2 }, we have v2 = a1 u1 + a2 u2 ,
where a1 = ⟨v2 , u1 ⟩ and a2 = ⟨v2 , u2 ⟩. Then
⟩
⟩
⟨
⟨
2
1
1
2
1
3
1
3
a1 = −1 , √ 0 = √
and a2 = −1 , √ −2 = √ .
2 1
2
6 −1
6
1
1
For v3 = b1 u1 + b2 u2 + b3 u3 , the coordinates are
⟩
⟩
⟨
⟨
1
1
1
1
√
1
6
1
6
b1 = 1 , √ 0 = √ ,
b2 = 1 , √ −2 = − √ = − 6,
2 1
2
6 −1
6
5
5
and
⟩
⟨
1
1
√
3
1
b3 = 1 , √ 1 = − √ = − 3.
3 −1
3
5
3
Then
√
2
[v2 ]B = √36 ,
0
B
√
3 2
√
[v3 ]B = − 6 .
√
− 3 B
These are easily checked.
The transition matrix from S coordinates to standard coordinates is given by the
matrix PS above. The transition matrix from B coordinates to standard
coordinates is given by the matrix
1
1
1
√
2
PB = 0
√1
2
√
6
−2
√
6
√
− 16
√
3
√1 .
3
− √13
CHAPTER 10.
108
Then the transition matrix from S coordinates to B coordinates is
P = PB−1 PS = PBT PS , since PB is an orthogonal matrix. We have,
√
√
√
1
√
√1
0
1
2
1
2
3/
2
3
2
2
2
√ √
√
√1
T
2
1
PB PS = 6 − √6 − √6 0 −1 1 = 0
3/ 2 − 6 .
√
√1
√1
− √13
1 1 5
0
0
− 3
3
3
Notice that the last column is the vector v3 expressed in B coordinates, so that,
indeed,
0
[v3 ]B = P [v3 ]S = P 0 .
1 S
Problem 10.8 An orthonormal basis {u1 , u2 , u3 } is obtained as follows:
1
√
1
2
0
0
=⇒
u1 =
v1 =
√1 .
1
2
0
0
1 1 5 1
√
3
3
⟩ √2
⟨ 3
2
2
2
0
0 0 0 0 0 0
w2 =
2 − 2 , √1 √1 = 2 − 5 = − 1
2
2
2
2
0
0
0
0
0
0
0
1
√
2
0
=⇒ u2 =
− √1 .
2
0
1 1
1 1
√
√
√
2
⟩ √2
⟩
⟨ 2
⟨ 2
2
2
2
1
1 0 0
1 0 0
w3 =
−1 − −1 , √1 √1 − −1 , − √1 − √1
2
2
2
2
3
0
0
0
0
3
3
1 3
0
2
2
2
1 0 0 1
=
−1 − 1 − − 3 = 0
2
2
3
3
0
0
109
=⇒
0
√1
10
u3 =
0
√3 .
10
Then {u1 , u2 , u3 } is an orthonormal basis of the subspace Lin{v1 , v2 , v3 }.
Problem 10.9
1 1 −1
A = −1 0 1
1 2 −1
2
1 0 −1 −1
1 → ··· → 0 1 0
3 .
5
0 0 0
0
The solutions of Ax = 0 are:
1
1
x1
s+t
x2 −3t
=
= s 0 + t −3 = sv1 + tv2 ,
x=
0
1
x3 s
1
0
x4
t
s, t ∈ R.
A basis of the null space is {v1 , v2 }. To obtain an orthonormal basis, let
1
1
0
.
u1 = √
2 1
0
Then set
1
1
1 ⟩
1
1
1
⟨ 1
2
−3
−3 1 0 1 0 −3 1 0 −3
w2 =
0 − 0 , √2 1 √2 1 = 0 − 2 1 = − 1 .
2
1
1
0
0
1
0
1
b 2 be the parallel vector w
b 2 = (1,√
For ease of calculation, let w
−6, −1, 2)T and
b 2 ⊥ u1 and that it is in N (A). We have ∥w
b 2 ∥ = 42. Then
check that w
1
1
1 1
0
−6
{u1 , u2 } = √ , √
1
42 −1
2
0
2
CHAPTER 10.
110
is an orthonormal basis of N (A).
The row space of A is orthogonal to the null space of A. A basis of the row space
of A is given by the non-zero rows of the reduced row echelon form of A, but
these vectors are not orthogonal. However, the first two rows of the matrix A also
form a basis of RS(A),
−1
1
0
1
{v3 , v4 } =
−1 , 1 ,
2
1
and these two vectors happen to be orthogonal, so we only need to normalise
them. Let u3 = √17 v3 and u4 = √13 v4 . Then {u1 , u2 , u3 , u4 } is a basis of R4
satisfying the requirements of the question. There are, of course, other possible
answers.
Problem 10.10 To find the line of intersection of the planes, solve the
homogeneous system of equations
x − y + 2z = 0
3x + 2y + z = 0.
using Gaussian elimination. We have
(
)
(
1 −1 2
1
A=
→ ··· →
3 2 1
0
0 1
1 −1
)
,
so the line of intersection of the planes is
−1
x
y =t
1
, t∈R .
S=
z
1
Therefore the vector (−1, 1, 1)T is in both planes.
To find the vectors x, y, z in R3 with the required properties, start with
−1
1
x = √ 1 .
3
1
111
Find a linearly independent vector in U . A natural choice is u = (1, 1, 0)T , and
luckily this vector satisfies ⟨x, u⟩ = 0, so the vectors are orthogonal. Then let
1
1
y = √ 1.
2 0
The set {x, y} is an orthonormal basis of U .
For V , a linearly independent vector is v = (0, 1, −2), for example. Apply the
Gram-Schmidt process. Set
⟨
⟩
1
0
−1
0
−1
0
−3
1
1
4
.
w=
1
−
1
,√
1
=
1
−
1
=
3
3
3
5
−2
−2
1
−2
1
−3
Check that the vector (−1, 4, −5)T is orthogonal to x and that it is in V . Then let
−1
1
z= √
4 .
42 −5
The set {x, z} is an orthonormal basis of V .
The set {x, y, z} is a basis of R3 . Since z ∈
/ Lin{x, y} = U , the three vectors are
linearly independent and three linearly independent vectors in R3 are a basis.
The set is not an orthonormal basis of R3 as ⟨y, z⟩ ̸= 0.
112
CHAPTER 10.
Chapter 11
Problem 11.1 The eigenvalues of the matrix A are λ1 = 8 and λ2 = −2.
Corresponding eigenvectors are non-zero scalar multiples of v1 = (1, 1)T and
v2 = (−1, 1)T . If
( 1
)
(
)
√
√1
−
8
0
2
2
P =
, D=
,
√1
√1
0 −2
2
2
then P T AP = D.
Problem 11.2 To show that v1 is an eigenvector, calculate the matrix product
Bv1 . Since
1 1 0
1
6
1
5 = 30 = 6 5 = 6v1 ,
Bv1 = 1 4 3
0 3 1
3
18
3
v1 is an eigenvector with corresponding eigenvalue λ = 6.
To find the other eigenvalues, solve the characteristic equation, |B − λI| = 0.
The determinant expansion by the first row will lead to an expression with a
common factor of (1 − λ), so it can be easily factorised to find the eigenvalues
are 1, −1, 6.
We already have an eigenvector corresponding to λ1 = 6. For λ2 = 1, solving
(A − I)x = 0, you should find that the eigenvectors are all non-zero scalar
multiples of v2 = (−3, 0, 1)T . Similarly for λ3 = −1, you obtain
113
CHAPTER 11.
114
v3 = (1, −2, 3)T . Note that these three eigenvectors are mutually orthogonal.
Taking the corresponding unit eigenvectors as the columns of a matrix Q, we
have that if
√
√
√
1/√35 −3/ 10 1/ √14
6 0 0
0
−2/√ 14
and
D = 0 1 0
Q = 5/√35
√
3/ 35 1/ 10
3/ 14
0 0 −1
then QT BQ = Q−1 BQ = D, and the matrix B is orthogonally diagonalised.
Problem 11.3 To find the eigenvalues, find the characteristic equation,
|A − λI| = 0. The determinant expansion by the first row will lead to an
expression with a common factor of (λ + 3), so it can be factorised to find the
eigenvalues are λ1 = 6 and λ2 = −3 of multiplicity two:
1−λ
−4
2
−4
2
1−λ
−2
= −(λ + 3)(λ + 3)(λ − 6) = 0.
−2
−2 − λ
To find the eigenvector for λ1 = 6, solve (A − 6I)v = 0,
−5 −4 2
1 0 −2
2
(A−6I) = −4 −5 −2 → · · · → 0 1 2 =⇒ v1 = −2 .
2 −2 −8
0 0 0
1
2
1
So u1 = −2 is an orthonormal basis of the eigenspace for λ = 6.
3
1
For λ = −3,
1 −1
4 −4 2
→
0 0
(A+3I) = −4 4 −2
2 −2 1
0 0
1
2
0
0
x
s − 12 t
=⇒ y = s
z
t
(The solutions are found by setting the free variables y = s and z = t.)
Two linearly independent eigenvectors are
1
−1
v2 = 1 ,
v3 = 0 .
0
2
115
Using the Gram-Schmidt process to obtain an orthonormal basis, let
1
√
2
u2 = √1 .
2
0
Then set
√1 1
⟨
√1 ⟩ √1
−1
−1
−1
−
2
2
1 12 1 2
1 √1
√
√
w=
0 −
=
=
0
,
0 +√
2
2
2
2
2
2
2
0
0
2
0
2
b be the parallel vector w
b = (−1, 1, 4)T , and check that it is orthogonal to
Let w
u2 , and to u1 . Then set
−1
1
u3 = √ 1 .
3 2
4
The set {u2 , u3 } is an orthonormal basis of the eigenspace for λ = −3, and the
set {u1 , u2 , u3 } is an orthonormal basis of R3 consisting of eigenvectors of A. If,
for example,
√1
2
− 3√1 2
−3 0 0
3
2
1
√
P = √12
− 23 ,
D = 0 −3 0 ,
3 2
4
1
√
0
0
0 6
3
3 2
Then P is an orthogonal matrix and P T AP = P −1 AP = D. This can be
checked by multiplying the matrices:
√1
√1
2
√1
0
1 −4 2
− 3√1 2
3
2
2
2
1
2
√
−
P T AP = − 3√1 2 3√1 2 3√4 2 −4 1 −2 √12
= D.
3
3 2
2
2
1
4
1
√
−3
2 −2 −2
0
3
3
3
3 2
Problem 11.4 (a) (Note that the problem should have referred to the vector v1
rather than v.) To show that the vector v1 is an eigenvector of A, mulitply Av1 .
3 −2 1
−1
−2
Av1 = −2 6 −2 0 = 0 = 2v1 ,
1 −2 3
1
2
116
CHAPTER 11.
which shows that v1 is an eigenvector of A with corresponding eigenvalue λ = 2.
You know that the matrix A can be diagonalised, even though there are only two
different eigenvalues, because the matrix A is symmetric, and any symmetric
matrix can be diagonalised.
You are given that λ = 8 is the only other eigenvalue of the matrix A. As there
are only two eigenvalues, one of them occurs with multiplicity two. First find the
eigenvectors for λ = 8, by solving (A − 8I)v = 0.
−5 −2 1
1
1
1
1 1
1
(A − 8I) = −2 −2 −2 −→ −5 −2 1 −→ 0 3
6
1 −2 −5
1 −2 −5
0 −3 −6
1 1 1
1 0 −1
−→ 0 1 2 −→ 0 1 2 .
0 0 0
0 0 0
1
1
So eigenvectors are v = t −2 , t ∈ R, t ̸= 0. Let v3 = −2 . There is
1
1
only one linearly independent eigenvector for λ = 8, and since you know that the
matrix can be diagonalised, there must be two linearly independent eigenvectors
for λ = 2. To find them, solve (A − 2I)v = 0.
1 −2 1
1 −2 1
(A − 2I) = −2 4 −2 −→ 0 0 0 .
0 0 0
1 −2 1
Set the non-leading variables equal to arbitrary parameters, y = s and z = t.
Then the solutions are
2s − t
2
−1
x = s = s 1 + t 0 , s, t ∈ R.
t
0
1
This method assures you that there are two linearly independent eigenvectors for
λ = 2, but you could have deduced them from what you already know. Since A
is symmetric, you know that eigenvectors corresponding to distinct eigenvalues
are orthogonal. Once you found the eigenvector for λ = 8 you knew that λ = 2
117
must be an eigenvalue of multiplicity two so its eigenspace must be a plane, and
this plane must be orthogonal to the eigenvector for λ = 8. So the normal of the
plane is the eigenvector for λ = 8, which is v3 = (1, −2, 1)T . You already have
one vector in the plane, namely v1 = (−1, 0, 1)T , so all you need to do is find
another linearly independent vector in the same plane, x − 2y + z = 0 and it can
be used as your choice for the second eigenvector; for example, v2 = (2, 1, 0)T .
Set
−1 2 1
P = 0 1 −2
1 0 1
and
2 0
D= 0 2
0 0
0
0.
8
Then P −1 AP = D. (You can write the eigenvectors as the columns of P in any
order, but the eigenvalues in the columns of D must match.)
Check that AP = P D to ensure that your eigenvectors are correct.
−2 4
8
−1 2 1
3 −2 1
AP = −2 6 −2 0 1 −2 = 0 2 −16 = P D
2 0
8
1 0 1
1 −2 3
↑ ↑ ↑
↑ ↑ ↑
v1 v2 v3
2v1 2v2 8v3
To orthogonally diagonalize the matrix A you need to find an orthogonal matrix
Q such that QT AQ = D. The columns of Q will be an orthonormal basis of R3
consisting of eigenvectors for A. The vector v3 is orthogonal to both v1 and v2 ,
but these vectors are not orthogonal to each other. So to obtain an orthonormal
basis of the eigenspace for λ = 2, you can use the Gram-Schmidt method.
⟨
⟩
2
2
−1
−1
−1
1
1
1
Let u1 = √ 0 . Set w = 1 − 1 , √ 0 √ 0 .
2 −1
2 −1
2 −1
0
0
Then
( ) −1
1
2
1
1
1
w = 1 +2
1 ,
0
= 1 . Set u2 = √
2
3
1
0
−1
1
1
1
u3 = √ −2 .
6
1
CHAPTER 11.
118
If
− √12
Q= 0
√1
2
√1
3
√1
3
√1
3
√1
6
− √26
√1
6
and
2 0
D = 0 2
0 0
0
0,
8
then QT AQ = D.
(An alternative solution is to take v1 = √15 (2, 1, 0)T and then apply the
Gram-Schmidt process, so that
1
−1
−5
( ) 2
1
1 = 2 .
w = 0 +2
5
5
1
1
0
−1
1
1
1
Set v2 = √ 2 and u3 = √ −2 .
30
6
5
1
Then
2
√
5
1
b
√
Q= 5
0
− √130
√2
30
√5
30
√1
6
− √26
√1
6
and
2
D = 0
0
0
0
2
0,
0
8
Other variations are possible.)
(b) In terms of x, y and z, the quadratic form f (x, y, z) = xT Ax (reading from
the matrix A) is given by
f (x, y, z) = xT Ax = 3x2 + 6y 2 + 3z 2 − 4xy + 2xz − 4yz.
Since we know the eigenvalues of A are λ = 2 and λ = 8, which are both
positive, the quadratic form is positive definite.
If B is the basis {u1 , u2 , u3 } obtained in part (a), corresponding to the columns
of Q, then
f (x, y, z) = xT Ax = zT Dz = 2X 2 + 2Y 2 + 8Z 2 ,
where x = Qz and z = (X, Y, Z)T . That is, X, Y, Z are coordinates in the basis
B. The matrix Q is the transition matrix from B coordinates to standard
coordinates.
119
Evaluating f (x, y, z) at a unit eigenvector u = (u1 , u2 , u3 )T for each eigenvalue
λ, you will find that f (u1 , u2 , u3 ) = λ. (See the next problem.) For example,
f ( √16 , − √26 , √16 ) = 3( 16 ) + 6( 46 ) + 3( 16 ) + 4( 62 ) + 2( 16 ) + 4( 26 ) = 8.
Problem 11.5 If A is a symmetric matrix and if u is a unit eigenvector
corresponding to the eigenvalue λ, then
uT Au = uT (Au) = uT (λu) = λ(uT u) = λ∥u∥2 = λ · 1 = λ.
Problem 11.6 Put the matrix A into reduced row echelon form
−5 1
2
1 0 − 21
A = 1 −5 2 → · · · → 0 1 − 12 .
2
2 −2
0 0 0
The solutions of Ax = 0 are
1
x = t 1 , t ∈ R,
2
so a basis of the null space consists of the vector w = (1, 1, 2)T .
The row space and column space are equal because the matrix A is symmetric, so
the row vectors and column vectors of the matrix are precisely the same set of
vectors. This space has dimension two, since rank(A) = 2. A two-dimensional
subspace of R3 is a plane.
Since the row space of A is orthogonal to the null space, the row space must be
the plane with normal vector w. This is also the column space of A. The
Cartesian equation is x + y + 2z = 0.
To show that v is an eigenvector, multiply Av:
−5 1
2
−1
6
−1
A = 1 −5 2 1 = −6 = −6 1 .
2
2 −2
0
0
0
So v is an eigenvector with corresponding eigenvalue λ = −6.
CHAPTER 11.
120
To find all the eigenvectors corresponding to λ = −6, solve (A + 6I)v = 0. We
have
1 1 2
1 1 2
(A + 6I) = 1 1 2 → 0 0 0 ,
2 2 4
0 0 0
with solutions
−1
−2
x = s 1 + t 0 ,
0
1
s, t ∈ R.
At this point you might notice that the eigenspace of λ = −6 is the plane with
normal vector (1, 1, 2)T .
Indeed, since the system of equations Ax = 0 has non-trivial solutions, λ = 0 is
an eigenvalue of A, and the null space is the eigenspace of λ = 0. Since A is
symmetric, the eigenspace for λ = 0 is orthogonal to the eigenspace for λ = −6.
So to orthogonally diagonalise A, we just need to find an orthonormal basis of
the eigenspace of λ = −6. Using the Gram-Schmidt process, let
u1 = √12 (−1, 1, 0)T . Then set
⟨
⟩
−2
−2
−1
−1
−1
1
1
v2 = 0 − 0 , √ 1 √ 1 = −1
2
2
1
1
1
0
0
Therefore if we take
√1
− 2 − √13
P = √12 − √13
√1
0
3
√1
6
√1 ,
6
√2
6
−6 0
D = 0 −6
0
0
0
0,
0
then P T AP = D.
The column space of A and the eigenspace of λ = −6 are the same space, the
plane in R3 with Cartesian equation x + y + 2z = 0. This happens for this
particular matrix A since the null space is the eigenspace of the eigenvalue of
multiplicity one, λ = 0. Looking at the subspaces of R3 as eigenspaces, the
eigenspace of λ = 0 is orthogonal to the eigenspace of λ = −6, which is a plane
since it has dimension two. At the same time, the null space is orthogonal to the
row space, which is equal to the column space. So the column space and the
eigenspace of λ = −6 are the same plane in R3 .
121
Problem 11.7 The quadratic form F (x, y) = 3x2 − 8xy + 3y 2 is equal to xT Ax,
where A is the symmetric matrix
(
)
3 −4
A=
.
−4 3
Since |A| = 9 − 16 < 0, the quadratic form is indefinite, and can assume positive
or negative values. For example, F (1, 0) = 3, and F (1, 1) = −2.
The quadratic form G(x, y) = 43x2 − 48xy + 57y 2 is equal to xT Bx, with
(
)
43 −24
B=
.
−24 57
Since |B| = (43)(57) − (24)2 > 0 and b11 = 43 > 0, the matrix B is positive
definite, and so the quadratic form is positive definite. Thus G(x, y) is never
negative, but, for example, G(1, 0) = 43 > 0.
Problem 11.8 (a) We can express f (x, y, z) = 6yz − x2 + 2xy − 4y 2 − 6z 2 as
xT Ax, where A is the symmetric matrix
−1 1
0
1 −4 3 .
0
3 −6
The principal minors of A are
−1 1
0
|A| = 1 −4 3 = −9 < 0,
0
3 −6
−1 1
= 3 > 0,
1 −4
−1 < 0,
so the quadratic form is negative definite.
(b) For the quadratic form g(x, y, z) = 6yz − x2 + 2xy − 4y 2 , we have
g(x, y, z) = xT Ax where A is the symmetric matrix
−1 1 0
1 −4 3 .
0
3 0
CHAPTER 11.
122
The principal minors of A are
|A| =
−1 1
1 −4
0
3
0
3 = −3(−3) > 0,
0
−1 1
= 3 > 0,
1 −4
−1 < 0,
so the quadratic form is indefinite.
Problem 11.9 If A is the symmetric matrix
3 2 −1
A = 2 2 1 ,
−1 1 5
then f (x, y, z) = 3x2 + 4xy + 2y 2 + 5z 2 − 2xz + 2yz = xT Ax. The principal
minors of A are
a11 = 3 > 0,
3
2
2
= 2 > 0,
2
|A| = 1 > 0,
so the quadratic form is positive definite. Therefore the eigenvalues of A are all
positive real numbers.
Problem 11.10 Let A be a positive definite n × n matrix. To show that
⟨x, y⟩ = xT Ay
defines an inner product on Rn , you need to show that the three properties are
satisfied.
Property (i) follows from the fact that A is symmetric:
⟨x, y⟩ = xT Ay = (xT Ay)T , since this is a 1 × 1 matrix. Then
⟨x, y⟩ = xT Ay = (xT Ay)T = yT AT x = yT Ax = ⟨y, x⟩.
Property (ii) is a direct result of the distributive properties of matrix
multiplication:
⟨αx + βy, z⟩ = (αx + βy)T Az = (αxT + βyT )Az
= αxT Az + βyT Az = α⟨x, z⟩ + β⟨y, z⟩.
Property (iii) follows immediately from the fact that A is positive definite:
⟨x, x⟩ = xT Ax ≥ 0 for all
x ∈ Rn
and
⟨x, x⟩ = 0 ⇐⇒ x = 0.
123
Problem 11.11 Given
1 −3
B = −1 3 ,
2
λ
you are asked to determine for which value(s) of λ the matrix B T B will be
invertible. There are two ways you can answer this question. You can state that
B T B is invertible if and only if B has full column rank, so when the column
vectors of B are linearly independent. (See Exercise 11.6.) Since there are only
two columns, this will be when they are not scalar multiples of one another, so
for all λ ̸= −6.
Alternatively, B T B is invertible if and only if |B T B| ̸= 0. Now,
(
(
)
)
1 −3
1 −1 2
6
−6
+
2λ
T
B B=
−1 3 =
,
−3 3 λ
−6 + 2λ 18 + λ2
2
λ
then B T B is not invertible if and only if
|B T B| = 108 + 6λ2 − 36 + 24λ − 4λ2 = 2λ2 + 24λ + 72 = 2(λ + 6)2 = 0.
So B T B is invertible for all λ ̸= −6.
For what value(s) of λ will the matrix B T B be positive definite? Again there are
two ways to answer this question. The matrix B T B will be positive definite if B
has full column rank, so provided λ ̸= −6. Alternatively, B T B is positive
definite if all its principal minors are positive. Since 6 > 0, and
|B T B| = 2(λ + 6)2 , this will also occur for all λ ̸= −6.
Problem 11.12 The first principal minor of any symmetric matrix is the (1, 1)
entry. For the matrix AT A, the (1, 1) entry is the inner product of the first row
vector of A with itself. Since ⟨x, x⟩ ≥ 0 for any vector x ∈ Rk , this scalar can
never be negative, so the matrix AT A can never be negative definite.
You might notice that unless A has a row of zeros, every entry on the main
diagonal of AT A is a strictly positive number.
CHAPTER 11.
124
Problem 11.13 The eigenvalues of the matrix
(
)
1 2
A=
2 1
are λ1 = 3 and λ2 = −1, with corresponding eigenvectors v1 = (1, 1)T and
v2 = (−1, 1)T . If
( 1
(
)
)
√
− √12
3 0
2
P =
and D =
, then P T AP = D.
√1
√1
0
−1
2
2
The linear transformation defined by P is a rotation anticlockwise by π/4
radians. Change coordinates by setting x = P z, with z = (X, Y )T . Then
xT Ax = (P z)T AP z = zT P T AP z = zT Dz.
The new X and Y axes are obtained by rotating the x and y axes by π4 radians
anticlockwise. In (X, Y ) coordinates, C is given by
zT Dz = 3X 2 − Y 2 = 3.
This is the equation of a hyperbola. It intersects the new X, Y axes in the points
(X, Y ) = (±1, 0).
First sketch the x and y axes, and then sketch the (rotated) X and Y axes. Then
mark the points where the curve C intersects the new axes. To sketch C you also
need the asymptotes (to see how wide or narrow the curves are).
The sketch is made easier by finding and using the points of intersection with the
old x and y axes. To find these, use the original equation,
xT Ax = x2 + 4xy + y 2 = 3.
√
√
If y = 0, we get (x, y) = (± 3, 0), and if x = 0 we get (x, y) = (0, ± 3).
√
The asymptotes of C in the new coordinates are Y = ± 3X, lines through the
origin with gradients of absolute value greater than 1 in the new coordinates. You
can add these to your sketch by noticing that the old y axis is the line Y = X
√ in
new coordinates and the old x axis is the line Y = −X, so the lines Y = ± 3X
125
are slightly steeper than these (in the new coordinates). You should now be able
to sketch the hyperbola.
Here is the sketch of x2 + 4xy + y 2 = 3 :
y
Y
X
1
−1
x
√
Problem 11.14 If C is the curve with equation 3x2 + 2 3 xy + 5y 2 = 6, and if
A is the symmetric matrix
√ )
(
3
3
A= √
,
3 5
then the curve C is given by xT Ax = 6.
You need to orthogonally diagonalise this matrix. The eigenvalues are found as
follows:
√
3
3
|A − λI| = √
= λ2 − 8λ + 12 = (λ − 6)(λ − 2).
3 5
So the eigenvalues are λ1 = 2 and λ2 = 6.
For λ1 = 2, solve (A − 2I)v = 0:
√ )
√ )
(
(
3
3
1
1
√
,
(A − 2I) =
→
0 0
3 3
( √ )
− 3
v1 =
.
1
CHAPTER 11.
126
For λ2 = 6, solve (A − 6I)v = 0:
√ )
√ )
(
(
−3
3
1
−(1/
3)
−→
,
(A − 6I) = √
3 −1
0
0
(
v2 =
√1
3
)
.
Check that ⟨v1 , v2 ⟩ = 0, and then make these into unit vectors,
)
(
( √ )
1 √1
1 − 3
u2 =
and u1 =
,
3
1
2
2
to obtain the orthonormal basis {u2 , u1 } for R2 , so that if we set
√ )
)
( 1
(
3
6 0
−
2
2
√
,
P =
and D =
3
1
0 2
2
2
then P −1 AP = P T AP = D, and P defines a rotation anticlockwise by π3
radians. To change coordinates, set x = P z with z = (X, Y )T .
In the new (X, Y ) coordinates, the curve C is given by
zT Dz = 6X 2 + 2Y 2 = 6.
√
This is an ellipse. It intersects the X axis at (±1, 0) and the Y axis at (0, ± 3).
To sketch the curve, begin by sketching the new X and Y axes, obtained by
rotating the x and y axes by π3 radians anticlockwise. Then mark off the points of
intersection of the ellipse on the new axes, and sketch the ellipse.
y
X
Y
√
3
1
x
−1
√
− 3
127
Compare this with Problem 7.11, in which you were told to change coordinates
by rotating the plane by π6 radians clockwise. This can be accomplished by using
a different orthogonal diagonalisation of the matrix A. If instead we take
( √3
)
1
2
2
√
Q=
,
− 12 23
b where
using the orthonormal basis {−u1 , u2 }, then Q−1 AQ = QT AQ = D
(
)
2 0
b
D=
.
0 6
Here Q represents rotation clockwise by π6 radians, and the curve C is given by
2X 2 + 6Y 2 = 6, or X 2 + 3Y 2 = 3 (where the new X and Y axes are obtained by
rotating the old x and y axes by π6 radians clockwise). This is how C was
sketched in Problem 7.11, The ellipse is the same; either rotation enables one to
sketch it.
Problem 11.15 In matrix form, the curve 3x2 + 4xy + 6y 2 = 14 is given by
xT Ax = 14 where A is the matrix
(
)
3 2
A=
.
2 6
The eigenvalues of A are λ1 = 7 and λ2 = 2. Corresponding eigenvectors are
v1 = (1, 2)T and v2 = (−2, 1)T . If
( 1
)
(
)
√
− √25
7 0
5
P =
and D =
, then P T AP = D.
√2
√1
0
2
5
5
To sketch the curve in the xy-plane, set x = P z to define new coordinates
z = (X, Y )T . Then xT Ax = zT Dz. The curve in the new coordinates is
7X 2 + 2Y 2 = 14.
√
This is
an
ellipse
which
intersects
the
X
axis
at
(±
2, 0) and the Y axis at
√
(0, ± 7).
The linear transformation defined by P is a rotation anticlockwise, but not by an
angle which we recognise. Therefore the images of the x and y axes under P are
CHAPTER 11.
128
found by looking at the images of the standard basis vectors under the linear
transformation defined by P . Thus the direction of the positive X axis is given
by v1 = (1, 2)T and the direction of the positive Y axis is given by the vector
v2 = (−2, 1)T . We now sketch the ellipse in standard position on the X and Y
axes.
Y
X
y
√
√
7
2
x
√
− 2
√
− 7
Chapter 12
Problem 12.1 The sum is not direct. Recall that if a sum is direct, then each
vector in the sum can be expressed in exactly one way as x + y where
x ∈ X, y ∈ Y . Here, we have
(
)
(1, 0, 1, −1)T = (1, 0, 1, 0)T − (0, 0, 0, 1)T + 0,
but also
(1, 0, 1, −1)T = 0 + (1, 0, 1, −1).
Since these are two different ways of expressing the vector (1, 0, 1, −1)T in the
form x + y, the sum is not direct. Alternatively, we may see that the sum is not
direct by observing that X ∩ Y ̸= {0}. In fact, from what’s written above, it’s
clear that (1, 0, 1, −1)T lies in X ∩ Y .
A basis for X + Y is
{(1, 0, 1, 0)T , (0, 0, 0, 1)T , (0, 1, 0, 0)T },
for instance.
Problem 12.2 Note that Y = R(A) where A is the matrix whose columns are
the vectors spanning Y . Then, Y ⊥ = R(A)⊥ = N (AT ). So we find N (AT ). We
solve AT x = 0 by finding the reduced row echelon form of AT ,
(
)
(
)
1 3 −1 1
1 0 −4 −2
T
A =
−→ · · · −→
,
1 4 0 2
0 1 1
1
129
CHAPTER 12.
130
with solutions
4s + 2t
4
2
−s − t
= s −1 + t −1 = sv1 + tv2
x=
s
1
0
t
0
1
s, t ∈ R.
The vectors v1 and v2 above form a basis of Y ⊥ .
Problem 12.3 To prove that (U + V )⊥ = U ⊥ ∩ V ⊥ , we need to show that
(U + V )⊥ ⊆ U ⊥ ∩ V ⊥ and U ⊥ ∩ V ⊥ ⊆ (U + V )⊥ .
Let a ∈ (U + V )⊥ . Then ⟨a, x⟩ = 0 for all x ∈ U + V . For any u ∈ U we have
u = u + 0 ∈ U + V , so ⟨a, u + 0⟩ = ⟨a, u⟩ = 0 for all u ∈ U and a ∈ U ⊥ . In
the same way, any v ∈ V can be written as v = 0 + v ∈ U + V and so
⟨a, 0 + v⟩ = ⟨a, v⟩ = 0 for all v ∈ V , which means that a ∈ V ⊥ . Since a is in
both U ⊥ and V ⊥ , a ∈ U ⊥ ∩ V ⊥ , and we have (U + V )⊥ ⊆ U ⊥ ∩ V ⊥ .
Now let a ∈ U ⊥ ∩ V ⊥ . Then ⟨a, u⟩ = 0 and ⟨a, v⟩ = 0 for all u ∈ U and all
v ∈ V . Then for any vector x = u + v ∈ U + V , we have
⟨a, u + v⟩ = ⟨a, u⟩ + ⟨a, v⟩ = 0 + 0 = 0
so that a ∈ (U + V )⊥ . This completes the proof.
Problem 12.4 The vectors u and w are linearly independent (since neither is a
scalar multiple of the other). Therefore they are a basis of R2 . Then any vector
x ∈ R2 can be written as a unique linear combination,
x = au + bw,
where au ∈ U and bw ∈ W . But this says that R2 is the direct sum of U and W .
(a) To find P , we need to find P (x) for any x ∈ R2 . We begin by expressing x as
a linear combination of u and w. We have
(
)
(
)
( ) (
)( )
−1
−3
x
−1 −3
a
x = au + bw = a
+b
, or
=
.
2
5
y
2
5
b
131
The image P (x) is the vector au. Solving the matrix equation for (a, b)T using
the inverse matrix, we obtain
( ) (
)( ) (
)
a
5
3
x
5x + 3y
=
=
.
b
−2 −1
y
−2x − y
Then the projection P is given by
(
) (
) (
)( )
−1
−5x − 3y
−5 −3
x
P (x) = au = (5x + 3y)
=
=
= Ax.
2
10x + 6y
10
6
y
You can check that the matrix A is idempotent, and that R(A) = R(P ) = U and
N (A) = N (P ) = W .
The image P (e1 ) = (−5, 10)T . (Note that this is not the vector in Lin{u} which
is closest to e1 , because the projection P is not an orthogonal projection.)
(b) You can find T in the same way as you found
( )P in part (a), using a vector
2
which is orthogonal to u; for example, v =
. Then
1
(
)
( )
( ) (
)( )
−1
2
x
−1 2
a
x = au + bw = a
+b
, or
=
.
2
1
y
2 1
b
The image T (x) is the vector au. Solving the matrix equation for (a, b)T using
the inverse matrix, we obtain
( )
(
)( )
1 −1 2
a
x
=
b
2 1
y
5
Then the orthogonal projection T is given by
(
)(
) ( 1
) ( 1
x − 25 y
−1
1
2
5
5
T (x) = au = − x + y
=
=
5
5
− 25 x + 54 y
− 25
2
)( )
x
= AT x.
4
y
5
− 25
The matrix AT is idempotent and symmetric (check this), so this is the
orthogonal projection. You could also obtain the matrix AT using
(
)
−1
B=
and AT = B(B T B)−1 B T .
2
The image T (e1 ) = ( 15 , − 25 )T . This is the vector in Lin{u} which is closest to e1 .
CHAPTER 12.
132
Problem 12.5 Suppose the n × n matrix A satisfies Ap+1 = Ap for some
constant p ≥ 1. Then
Ap+2 = AAp+1 = AAp = Ap+1 = Ap .
Continuing in this manner, we have Ap+r = Ap , for all r ∈ N. If j > p, then
j − p = r ∈ N, so that Aj = Ap+r = Ap .
To show that Rn = R(Ap ) ⊕ N (Ap ), we can use an analogous argument to the
one given in the proof of Theorem 12.26. Let x ∈ Rn . We can write x as
x = Ap x + (x − Ap x).
Then Ap x ∈ R(Ap ), and since
Ap (x − Ap x) = Ap x − Ap+p x = Ap x − Ap x = 0,
we have (x − Ap x) ∈ N (Ap ). Therefore, Rn = R(Ap ) + N (Ap ).
To show the sum is direct, we show that R(Ap ) ∩ N (Ap ) = {0}. Indeed, let
z ∈ R(Ap ) ∩ N (Ap ). Then z ∈ R(Ap ) so there is a vector y ∈ Rn such that
Ap y = z. But also z ∈ N (Ap ), so Ap z = 0. Therefore,
0 = Ap z = Ap Ap y = Ap+p y = Ap y = z, so z = 0; that is, the only vector in
R(Ap ) ∩ N (Ap ) is the zero vector.
Problem 12.6 The subspace S = Lin{x} may be thought of as the range, R(x),
of the n × 1 matrix x (since each is the set of all linear combinations of x). We
know that the orthogonal projection of Rn onto R(A) for a general matrix A is
A(AT A)−1 AT , so the matrix we want is
(
)−1 T
1 T
P = x xT x
x = x(∥x∥2 )−1 xT = x
x .
∥x∥2
So P = (1/∥x∥2 )xxT .
Alternatively, it may easily be checked that P is symmetric and idempotent.
From this, it follows that P is the orthogonal projection onto R(P ). To show
R(P ) = Lin{x}, we first show that R(P ) ⊆ Lin{x}. If y ∈ Rn , we have
(
)
1 T
1
T
xx y = x
x y ,
Py =
∥x∥2
∥x∥2
133
which is a scalar multiple of x. Next, suppose z = αx ∈ Lin{x}. Then
(
)
1
1
1
T
T
T
xx (z) =
xx (αx) = α
xx x = αx = z,
∥x∥2
∥x∥2
∥x∥2
so Lin{x} ⊆ R(P ). This all shows that P is the orthogonal projection of Rn
onto Lin{x}.
Problem 12.7 The orthogonal projection of Rn onto R(A) for a general matrix
A is A(AT A)−1 AT . Taking A = z, R(A) = Lin{z}, so AT A = zT z = 18 and
the matrix we want is
2
4 −6 4 −2
−3 1
1
−6 9 −6 3 .
(
2
−3
2
−1
)
=
2 18
18 4 −6 4 −2
−1
−2 3 −2 1
(Alternatively, you could use the result of the previous problem.)
Problem 12.8 The two vectors are linearly independent, so they form a basis of
the linear span of the vectors, which is a plane in R3 . A Cartesian equation of the
plane is given by
0 2 x
1 1 y = −2x + 2y − 2z = 0,
1 −1 z
or x − y + z = 0. A normal vector to the plane is n = (1, −1, 1)T , so n is a
basis of X ⊥ .
The matrix representing the orthogonal projection onto X is given by
P = A(AT A)−1 AT where A is the matrix whose columns are the two given
vectors. We calculate this:
(
) 0 2
(
)
(
)
1
0 1 1
2
0
3
0
T
T
−1
A A=
1 1 =
(A A) =
2 1 −1
0 6
6 0 1
1 −1
Then
(
0 2
1
3
P =
1 1
0
6
1 −1
0
1
)(
0
2
1 1
1 −1
)
2 1 −1
1
= 1 2 1 .
3
−1 1 2
134
CHAPTER 12.
Since P is symmetric, it is diagonalisable.
The eigenspace corresponding to λ = 0 is N (P ), so n = (1, −1, 1)T is an
eigenvector for eigenvalue 0. The eigenspace corresponding to λ = 1 is R(P ),
so, for example, (0, 1, 1)T is an eigenvector for eigenvalue 1. These can also be
found directly by the usual method of finding eigenvectors.
Problem 12.9 We need to show that P 2 = P .
1
2 −2 0
1
2 −2 0
1
7 −1 3
7 −1 3
2
1 2
P2 =
9 −2 −1 7 3 9 −2 −1 7 3
0
3
3 3
0
3
3 3
9
18 −18 0
( )( )
1
1 18
63 −9 27
= P.
=
9
9 −18 −9 63 27
0
27
27 27
Since the matrix P is idempotent and symmetric, it represents an orthogonal
projection of R4 onto Y = R(P ).
For a subspace Y of a finite dimensional vector space V , such as R4 , the
orthogonal projection of V onto Y is the projection onto Y parallel to Y ⊥ . (That
is, P : V = Y ⊕ Y ⊥ → Y , with P (Y ) = Y and P (Y ⊥ ) = 0.)
We have Y = R(P ) and Y ⊥ = N (P ). To find these, put the matrix P into
reduced row echelon form.
1
2 −2 0
1 2 −2 0
1 0 −4 −2
1 2
7 −1 3
1
→ 0 3 3 3 → 0 1 1
.
P =
0 3 3 3
0 0 0
0
9 −2 −1 7 3
0
3
3 3
0 3 3 3
0 0 0
0
The positions of the leading ones in the reduced row echelon form indicate that a
basis of R(P ) is given by the first two columns of P . Therefore, a basis of
Y = R(P ) is
2
1
7
2
.
,
−2 −1
3
0
135
Alternatively, since P is symmetric, and hence the row space of P is the same as
R(P ), the column space of P , another basis is
1
0
1
0
−4 , 1 .
1
−2
To find a basis of the null space of P , we solve the system P x = 0, with
x = (x, y, z, w)T . Setting the non-leading variables equal to arbitrary
parameters, z = s and y = t, and reading the solutions from the reduced row
echelon form of P , we have
4
2
x
4s + 2t
y −s − t
= s −1 + t −1 .
=
0
1
z s
1
0
t
w
Then a basis of Y ⊥ = N (P ) is
4
2
−1
−1
1 , 0
0
1
Problem 12.10 Suppose A is an n × n diagonalisable matrix and that the only
eigenvalues of A are 0 and 1. Then there is an invertible matrix P and a diagonal
matrix D = diag(1, . . . , 1, 0, . . . , 0) such that P −1 AP = D. Since the only
entries of D are 0 and 1, D2 = D. We can write A = P DP −1 . Then
A2 = P DP −1 P DP −1 = P D2 P −1 = P DP −1 = A,
so A is idempotent.
Since A is idempotent, the linear transformation defined by T (x) = Ax is a
projection from Rn onto R(T ) = R(A) parallel to N (T ) = N (A). By Exercise
12.4, you know that R(A) is equal to the eigenspace corresponding to the
eigenvalue λ = 1. The null space, N (A) is the same as the eigenspace
corresponding to λ = 0.
CHAPTER 12.
136
Problem 12.11 The vector p is not on the plane because its components do not
satisfy the equation. The nearest point in U to p is the orthogonal projection of p
onto U . This can be found either by using the method of least squares, or directly
by finding the point of intersection of the plane and the line through p which is
perpendicular to the plane. We will show both.
Using lines and planes, the normal vector is n = (2, −1, −2)T , so the vector
equation of the line in R3 which passes through the point p and is parallel to n is
x
1
2
1 + 2t
x = y = 1 + t −1 = 1 − t .
z
2
−2
2 − 2t
Substituting the values of x, y, z into the equation of the plane, we have
2(1 + 2t) − (1 − t) − 2(2 − 2t) = 0.
Solving for t, we obtain t = 1/3. From this we deduce that the required point is
2
5/3
5
1
1
1
1 + −1 = 2/3 = 2 .
3
3
−2
4/3
4
2
Using the method of least squares, the projection of R3 onto U is given by the
matrix P = A(AT A)−1 AT , where A is a matrix whose columns are basis vectors
of the plane. These can be any two linearly independent vectors which satisfy the
Cartesian equation of the plane. For example,
(
)
(
)
1 1
1
5 1
2 −1
T
T
−1
Then A A =
, (A A) =
A= 2 0 .
.
1 2
9 −1 5
0 1
Calculating the matrix product,
(
)(
1 1
1
2
−1
1
P = 2 0
1
9 −1 5
0 1
Then P p is,
2
0
0
1
)
5 2
4
1
= 2 8 −2 .
9
4 −2 5
5 2
4
1
15
5
1
1 1
2 8 −2
1 =
6
=
2 ,
9
9
3
4 −2 5
2
12
4
137
which is the nearest point in U to p.
Problem 12.12 The equations are
2a + b
4a + b
5a + b
6a + b
=
=
=
=
13
17
22
25.
In matrix form, Ax = b where
13
2 1
( )
4 1
, x = a , b = 17 .
A=
22
5 1
b
25
6 1
The least squares solution is x∗ = (AT A)−1 AT b. This gives
( ∗) (
)
a
107/35
∗
x =
=
,
b∗
219/35
so we take a = 107/35 and b = 219/35. (The detailed calculations are omitted
here.)
Problem 12.13 In matrix form, the equations yi = mxi + c are
y1
x1 1
( )
x2 1 m
y2
.
=
..
..
... ,
. c
xn 1
yn
which is of the form Ax = b with x = (m, c)T . Then a least squares solution is
x∗ = (AT A)−1 AT b. Now,
x1 1
(∑ 2 ∑ )
)
(
x2 1
x
xi
x1 x2 · · · xn
T
,
A A=
..
.. = ∑ i
xi
n
1 1 ··· 1
.
.
xn 1
CHAPTER 12.
138
where
∑
denotes
T
∑n
i=1 . So,
−1
(A A)
1
= ∑ 2
∑
n xi − ( xi )2
(
−
n
∑
xi
∑ )
−∑ xi
.
x2i
Then,
x∗ = (AT A)−1 AT b
1
=
∑ 2
∑
n xi − ( xi )2
(
−
n
∑
xi
∑ )(
−∑ xi
x1
2
xi
1
x2
1
(
)
∑ )(∑
1
n
−∑ xi
xi y i
∑
∑
=
∑
∑
x2i
yi
n x2i − ( xi )2 − xi
(
)
∑
∑ ∑
1
n
x
y
−
x
y
i
i
i
i
∑ ∑
∑ 2∑
=
,
∑
∑
yi
n x2i − ( xi )2 − xi xi yi + xi
and, therefore,
y1
)
y
· · · xn
.2
· · · 1 ..
yn
∑
∑ ∑
xi yi − xi yi
,
m =
∑
∑
n x2i − ( xi )2
∑ ∑ 2 ∑ ∑
y i xi − xi xi y i
∗
c =
.
∑
∑
n x2i − ( xi )2
∗
n
(You can use this to obtain the solution to the previous problem.)
Chapter 13
√
Problem 13.1 Plot the points, as in R2 : z = ( 3, −1) and w = (1, 1). This will
enable you to deduce the modulus and argument of each:
z=
√
3 − i,
w = 1 + i,
so
so |z| = 2, θ = −
|w| =
√
2, θ =
π
4
π
6
=⇒ z = 2e−i 6 ,
π
=⇒ 1 + i =
√
2eiπ/4 .
Then,
( −iπ/6 )6
√
2e
( 3 − i)6
26 e−iπ
q=
=
=
= 2ei(−π−5π/2) = 2eiπ/2 = 2i.
(√
)
10
5 ei10/4
iπ/4
(1 + i)10
2
2e
Problem 13.2 To write each of the following complex numbers in the form
a + ib use Euler’s formula, so reiθ = r(cos θ + i sin θ).
e−i3π/2 = i,
√
3
1
,
e1+i = e1 ei = e cos(1) + ie sin(1).
=e
= −i
2
2
−1
e
is already in the form a + ib since it is a real number.
√
√
√
√
e−3+i2 5 = e−3 ei2 5 = e−3 cos(2 5) + ie−3 sin(2 5),
√
4ei7π/6 = −2 3 − 2i.
i(11π/3)
e
1
1
ei3π/4 = − √ + i √ ,
2
2
−i(π/3)
139
CHAPTER 13.
140
Problem 13.3 Find the roots w and w of x2 − 4x + 7 = 0 using the quadratic
formula:
√
√
√
4 ± −12
x=
=⇒ w = 2 + i 3 , w = 2 − i 3 are the solutions.
2
For this value of w, the real and imaginary parts of f (t) = ewt , t ∈ R are
obtained from the expression
√
√
√
√
ewt = e(2+i 3)t = e2t ei 3t = e2t (cos 3t + i sin 3t).
√
√
Then Re(f (t)) = e2t cos 3t, and Im(f (t)) = e2t sin 3t.
For w, the real part of f (t) is the same and the imaginary part differs only in
sign. We have
√
√
√
√
ewt = e(2−i 3)t = e2t e−i 3t = e2t (cos 3t − i sin 3t).
√
√
so that Re(f (t)) = e2t cos 3t, and Im(f (t)) = −e2t sin 3t.
To find the real and imaginary parts of g(t) = wt , t ∈ Z+ , we have
√
√
√
wt = (2 + i 3)t = ( 7eiθ )t = ( 7)t (cos θt + i sin θt),
√
3
2
where cos θ = √
and sin θ = √ .
7
7
Then Re(g(t)) = 7t/2 cos θt and Im(g(t)) = 7t/2 sin θt.
For w, the real part of g(t) = wt is the same and the imaginary part differs only
in sign. We have Re(g(t)) = 7t/2 cos θt and Im(g(t)) = −7t/2 sin θt.
Problem 13.4 For λ = a + ib we have
y(t) = Aeλt +Beλt = Ae(a+ib)t +Be(a−ib)t = Aeat eibt +Beat e−ibt = eat (Aeibt +Be−ibt ).
Using Euler’s formula, this becomes
y(t) = eat (A(cos bt + i sin bt) + B(cos bt − i sin bt))
= eat ((A + B) cos bt + i(A − B) sin bt)
b cos bt + B
b sin bt).
= eat (A
141
By choosing A and B to be complex conjugates for any complex number, say
b = A + B = 2Re(Z) and
A = Z and B = Z, we have A
b = i(A − B) = −2Im(Z), which are both real numbers. This makes y(t) a
B
real-valued function of t (as opposed to a complex-valued function).
b and B
b are real, this last form can be expressed as Ceat cos(bt − ϕ) using the
If A
formula for the cosine of the sum of two angles:
Ceat cos(bt − ϕ) = Ceat ( cos(bt) cos ϕ + sin(bt) sin ϕ)
b and sin ϕ = B/C,
b
b is the adjacent side of a right
with cos ϕ = A/C
so that A
b is the opposite side and C is the hypotenuse. Then
triangle, B
√
√
√
b2 + B
b 2 = (A + B)2 + i2 (A − B)2 = 2 AB.
C= A
Problem 13.5 Recall that eiθ and e−iθ are complex conjugates for any θ ∈ R, so
that in general
eiθ + e−iθ = (cos θ + i sin θ) + (cos θ − i sin θ) = 2 cos θ
To show that ezt + ezt is real for t ∈ R, write z = a + ib ∈ C,
ezt + ezt = e(a+ib)t + e(a−ib)t = eat (eibt + e−ibt ) = eat (2 cos bt)
which is real.
To show that z t + z t is real for t ∈ Z+ , write z = reiθ ,
z t + z t = rt eiθt + rt e−iθt = rt (eiθt + e−iθt ) = rt (2 cos θt).
Problem 13.6 The three roots of z 3 = −1 are of the form
z = reiθ ,
π
where r3 = 1,
3θ = π + 2nπ.
5π
The roots are: ei 3 , eiπ = −1, ei 3 .
Mark the point (−1, 0) in the complex plane. The other two roots (which are
complex conjugates) are on the unit circle, making angles of ± π3 with the
positive real axis.
CHAPTER 13.
142
The roots of z 4 = −1 were found in Exercise 13.1. They are:
π
3π
5π
7π
ei 4 , e i 4 , e i 4 , e i 4 .
To illustrate the roots as points in the complex plane, plot the first root in terms of
its polar coordinates, r = 1, θ = π/4. (If you have trouble with this, change the
first root to the form a + ib: eiπ/4 = √12 + i √12 and mark the point ( √12 , √12 ) on
the graph.) The remaining 3 points are on the unit circle, spaced evenly a
distance π2 apart.
For x5 = −1 note that −1 is a root and the remaining 4 roots will be evenly
spaced around the unit circle. For x6 = 64, 2 is a real root and the 6 roots are
evenly spaced about the circle of radius 2.
Problem 13.7 (a) To find the eigenvalues of the matrix C, we use the
characteristic equation,
|C − λI| =
1−λ
−1
= λ2 − 2λ + 2 = 0,
1
1−λ
with solutions λ = 1 ± i The eigenvalues of C are λ = 1 + i and λ = 1 − i.
To diagonalise C we find an eigenvector for each eigenvalue. For λ = 1 + i,
(
)
(
)
( )
−i −1
1 −i
i
C − (1 + i)I =
−→
.
Let v1 =
.
1 −i
0 0
1
The eigenvalue λ is the complex conjugate of λ, so the eigenvector
corresponding to λ is the complex conjugate of v1 . Therefore,
v2 = v1 = (−i, 1)T is an eigenvector for λ = 1 − i. Set
(
)
(
)
i −i
1+i
0
P =
and D =
.
1 1
0
1−i
Then P −1 CP = D. (Check CP = P D.)
(b) To solve y′ = Cy, set y = P z. Then y′ = Cy becomes the system z′ = Dz
with solutions
(
) ( λt )
z1 (t)
αe
, with α = z1 (0) and β = z2 (0).
z=
=
βeλt
z2 (t)
143
You can either find the constants α and β now using z(0) = P −1 y(0), or do this
later, which is the method we will use. Using y = P z,
(
) (
) ( λt ) (
)
y1 (t)
i −i
αe
αieλt − βieλt
=
=
.
βeλt
y2 (t)
1 1
αeλt + βeλt
y1 (t) = αieλt − βieλt
y2 (t) = αeλt + βeλt
We now use the initial conditions, y1 (0) = 0, y2 (0) = 1, to find α and β.
}
y1 (0) = 0 = αi − βi
1
=⇒ α = β = .
2
y2 (0) = 1 = α + β
So the general complex solution is
)
1 ( λt
λt
y1 (t) = i e − e
2
)
1 ( λt
λt
y2 (t) =
e +e
2
with λ = 1 + i
Simplifying these solutions using Euler’s formula, we have
eλt = e(1+i)t = et eit = et (cos t+i sin t)
eλt = e(1−i)t = et e−it = et (cos t−i sin t)
so that
(eλt − eλt ) = 2iet sin t
(eλt + eλt ) = 2et cos t.
The real solutions of y′ = Cy are therefore:
y1 (t) = −et sin t
y2 (t) = et cos t
.
Problem 13.8 The vector x is in U ∩ W if there are constants s, t, r, q ∈ C such
that
x
1
0
0
1
v = y = s 0 + t −i = r
i
+q
0 .
z
i
−1
−1
−i
CHAPTER 13.
144
This gives rise to the three equations s = q, −it = ir and is − t = −r − iq.
Substituting s = q and t = −r from the first two equations into the third, we
obtain is = t. One solution is s = 1, t = i, q = 1, r = −i, so that
1
0
0
1
1
v = s 0 +t −i = r i +q 0 = 1 is a basis of U ∩ W .
i
−1
−1
−i
0
1
1
To find an orthonormal basis of each set as instructed, first take x = √
1 .
2 0
Next take any other (linearly independent) vector in U , such as (1, 0, i)T and
apply Gram-Schmidt. Set
⟨
⟩
1 1
1
1
1
1
1
2
2
1
1
w = 0 − 0 , √ 1 √ 1 = 0 − 12 = − 12
2
2
i
i
0
0
i
0
i
1
1
Let y = √
−1 . Then {x, y} is an orthonormal basis of U .
6
2i
(Check that y is orthogonal to x and that it is in U .)
Now do the same for W . For example, the vector (1, 0, −i)T is a vector in W
which is linearly independent from x. We apply Gram-Schmidt to obtain
⟨
⟩
1 1
1
1
1
1
1
1 1 21 2 1
√
w=
0 −
0
,√
1
1 =
0 − 2 = −2 .
2 0
2 0
−i
−i
−i
0
−i
1
1
Let z = √ −1 . Then {x, z} is an orthonormal basis of W .
6 −2i
(Check that z is orthogonal to x and that it is in W .)
The set {x, y, z} is a basis of C3 . The vector z is not in U , since it is in W and it
is not in U ∩ W = Lin{x}, so the three vectors are linearly independent, and
145
three linearly independent vectors in C3 are a basis. It is not an orthonormal basis
of C3 as ⟨y, z⟩ =
̸ 0:
⟩
⟨
1
1
−1 −1 = (1)(1) + (−1)(−1) + (2i)(2i) = 2 − 4 ̸= 0.
2i
−2i
Problem 13.9 The standard inner product is the complex number given by
⟨x, y⟩ = x1 y1 + x2 y2 + x3 y3 . For the given vectors, we have
⟩
⟨
1
0
i , 2 − i = 0 + i(2 + i) + 1(1 − 2i) = 0
1
1 + 2i
so the vectors are orthogonal. An orthonormal basis of Lin{v1 , v2 } is {u1 , u2 }
where
1
0
1
1
u2 = √ 2 − i .
u1 = √ i
3 1
10 1 + 2i
To extend this to an orthonormal basis of C3 , we need to find a vector v3 which is
orthogonal to both v1 and v2 . This can be done by setting v3 = (a, b, c)T and
simultaneously solving the linear equations given by
⟨v3 , v1 ⟩ = 0
⟨v3 , v2 ⟩ = 0,
or by applying the Gram-Schmidt orthogonalisation process. The linear
equations are
{
a − ib + c = 0
−1 + 2i
=⇒ b =
c = ic, a = −2c.
2+i
(2 + i)b + (1 − 2i)c = 0
Letting c = 1 will give a solution, v3 = (−2, i, 1)T which can then be
normalised.
Alternatively, for the Gram-Schmidt process we can choose e1 = (1, 0, 0)T , and
let w3 be the vector
⟩
⟨
⟩
⟨
1
1
1
1
0
0
1
0 − 0 , √1 i √1 i − 0 , √1 2 − i √1 2 − i ,
3 1
3 1
10 1 + 2i
10 1 + 2i
0
0
0
CHAPTER 13.
146
1
2
1
1
1
w3 = 0 − i = −i .
3
3
1
−1
0
(At this point, check that w3 is orthogonal to v1 and v2 . Notice that the vector e1
was chosen to simplify the calculation.) Now set
−2
1
u3 = √ i .
6
1
so
Then {u1 , u2 , u3 } is an orthonormal basis of C3 .
To find the coordinates of (1, 1, i)T in this basis, calculate its inner product with
the basis vectors (in the correct order),
⟩
⟨
1
1
1
1
a1 =
=√
1 ,√
i
3 1
3
i
⟩
⟨
0
1
4 + 2i
1
a2 = 1 , √ 2 − i = √
10 1 + 2i
10
i
⟩
⟨
1
−2
1
2
a3 = 1 , √ i = − √
6
6
i
1
Then a = a1 u1 + a2 u2 + a3 u3 (which can be easily checked).
The matrix E1 is u1 u∗1 . We have
1
1
E1 = i ( 1 −i
3
1
1 −i
1
1) = i 1
3
1 −i
1
i .
1
A matrix A is Hermitian if A∗ = A, where A∗ denotes the Hermitian conjugate
of A. A matrix A is idempotent if A2 = A. For E1 ,
1 −i 1
1
E1∗ = i 1 i = E1
3
1 −i 1
147
so E1 is Hermitian, and E1 is idempotent since
1 −i 1
1 −i 1
3
1
1
E12 = E1 E1 = i 1 i i 1 i = 3i
9
9
1 −i 1
1 −i 1
3
−3i
3
−3i
3
3i = E1 .
3
You should find, by matrix multiplication, that E1 a = a1 u1 .
In the same way,
0
1
∗
E2 = u2 u2 =
0
10
0
0
5
5i
0
−5i
5
4
1
∗
E3 = u3 u3 =
−2i
6
−2
2i
1
−i
−2
i .
1
The matrices E1 , E2 and E3 are the matrices of the orthogonal projections of C3
onto the subspaces Lin{u1 }, Lin{u2 } and Lin{u3 } respectively. Then
I = E1 + E2 + E3 .
(Now you can reward your labours by checking that E1 + E2 + E3 = I!)
Problem 13.10 We look for the eigenvalues and corresponding eigenvectors of
the matrix
2
1+i 0
3
0.
A = 1 − i
0
0
5
The eigenvalues are found by solving |A − λI| = 0. Expand the determinant by
the third row to obtain
2−λ 1+i
1−i 3−λ
0
0
0
0
= (5 − λ)(λ2 − 5λ + 4) = (5 − λ)(λ − 1)(λ − 4) = 0,
5−λ
so the eigenvalues are λ1 = 5, λ2 = 1, λ3 = 4.
From the form of A, it is clear that a unit eigenvector for λ1 = 5 is
u1 = (0, 0, 1)T .
For λ2 = 1, we solve (A − I)v = 0:
1
1+i 0
1
A−I = 1−i
2
0 −→ 0
0
0
4
0
1+i
0
0
0
1
0
CHAPTER 13.
148
from which we deduce that (1 + i, −1, 0)T is an eigenvector. A unit eigenvector
is u2 = √13 (1 + i, −1, 0)T .
An eigenvector for λ = 4 is found in a similar way:
−2 1 + i 0
1 −(1 + i)/2 0
A − 4I = 1 − i −1 0 −→ 0
0
1
0
0
1
0
0
0
from which we deduce that (1 + i, 2, 0)T is an eigenvector. Let
u3 = √16 (1 + i, 2, 0)T .
An orthonormal basis of C3 consisting of eigenvectors of A is given by
{u1 , u2 , u3 } where
0
1+i
1+i
1
1
u1 = 0
u2 = √ −1
u3 = √ 2 .
3
6
1
0
0
The matrix A is Hermitian, since A∗ = A (and therefore, as expected, the
eigenvalues of A are real numbers). Since A is Hermitian, it is also normal.
T
Problem 13.11 If A is a complex matrix, then A∗ = A is the Hermitian
conjugate (the complex conjugate transpose) of A. The matrix A is normal if
AA∗ = A∗ A.
If the matrix A is real, then this becomes AAT = AT A. Applying this to the
2 × 2 matrix, we have
(
)(
) (
)(
)
a b
a c
a c
a b
=
c d
b d
b d
c d
or
(
a2 + b2 ac + bd
ac + bd c2 + d2
)
(
=
a2 + c2
ab + cd
ab + cd
b2 + d2
)
,
which leads to the equations
b2 = c2
and
ac + bd = ab + cd.
The first equation tells us that b = c or b = −c. If b = c, then both equations are
satisfied and the matrix A is symmetric, A = AT .
149
If b = −c, then the second equation yields ac = cd, so either c = 0, in which
case A is symmetric (because it is diagonal), or a = d, in which case
(
)
a b
,
A=
−b a
which is a multiple of an orthogonal matrix, since the column vectors are
orthogonal and of the same length, but not necessarily unit vectors.
Problem 13.12 The matrix A is not Hermitian, as it is real and it is not
symmetric, AT ̸= A. However, A is normal. You can show this directly by
calculating that A∗ A = AA∗ . Note that this matrix A is a scalar multiple of an
orthogonal matrix (see the previous problem) since its columns are orthogonal
vectors of the same length.
The eigenvalues of the matrix
)
2 1
−1 2
are found by solving the characteristic equation,
2−λ
−1
(
1
= λ2 − 4λ + 5 = 0,
2−λ
with solutions λ = 2 ± i. Solving (A − λI)v = 0 for λ = 2 + i, we find
(
)
(
)
( )
−i 1
1 i
−i
A − (2 + i)I =
→
, so v1 =
−1 −i
0 0
1
is a corresponding eigenvector.
Since the matrix A is real, we have that the other eigenvalue is the complex
conjugate, λ = 2 − i, with corresponding eigenvector v2 = v1 .
Because A is normal, these vectors will be orthogonal. This can also be checked
directly:
⟨( ) ( )⟩
i
−i
,
= (i)(i) + 1 = 0.
1
1
So to unitarily diagonalise A, we only need to normalise these, obtaining
u1 = √12 v1 and u2 = √12 v2 . Then, for
(
)
(
)
1
i −i
2−i
0
P =√
and D =
,
0
2+i
2 1 1
CHAPTER 13.
150
we have P ∗ AP = D.
The spectral decomposition of A is A = (2 + i)u1 u∗1 + (2 − i)u2 u∗2 , which is
( )
( )
2−i i
2 + i −i
A=
(i 1) +
( −i 1 ) .
1
1
2
2
Multiplying out the matrices, the spectral decomposition is
)
)
(
(
2−i 1 i
2 + i 1 −i
+
.
A=
i 1
−i 1
2
2
Problem 13.13 If the matrix A is unitary, its columns must be an orthonormal
basis of Cn , so the only possibility is z = 0. The matrix
(
)
0 i
A=
1 0
is unitary; its columns are an orthonormal basis of C3 .
(Alternatively, if A is unitary, then AA∗ = A∗ A = I, and you can show AA∗ = I
for the given matrix A if and only if z = 0.)
For the matrix A to be normal, we need AA∗ = A∗ A, where
(
)(
) (
)
0 i
0 1
1
iz
∗
AA =
=
1 z
−i z
−iz 1 + zz
(
)(
) (
)
0 1
0 i
1
z
A A=
=
.
−i z
1 z
z 1 + zz
These matrices will be equal if and only if z = iz. If z = a + ib, then we need to
have
a + ib = i(a − ib) = ia + b,
and
∗
so that a = b. The matrix A is therefore normal if and only if
z = a + ia = a(1 + i), for some a ∈ R. (You should check this.)
Problem 13.14 To unitarily diagonalise the matrix A
2 i 0
A = −i 2 0
0 0 5i
151
we first find the eigenvalues and corresponding eigenvectors. The eigenvalues are
found by solving |A − λI| = 0. Expand the determinant by the third row to
obtain the characteristic equation
2−λ
i
−i
2−λ
0
0
0
= (5i − λ)(λ2 − 4λ + 3) = (5i − λ)(λ − 1)(λ − 3) = 0.
0
5i − λ
So the eigenvalues are λ1 = 5i, λ2 = 1, λ3 = 3.
From the form of A, it is clear that a unit eigenvector for λ1 = 5i is
u1 = (0, 0, 1)T . For λ2 = 1, we solve (A − I)v = 0:
1 i
0
1 i 0
A − I = −i 1
0 → ··· → 0 0 1,
0 0 5i − 1
0 0 0
from which we deduce that (−i, 1, 0)T is an eigenvector. A unit eigenvector is
u2 = √12 (−i, 1, 0)T .
The eigenvector for λ = 3 is found in the same way:
1 −i
−1 i
0
→ ··· → 0 0
A − 3I = −i −1
0
0 0
0
0 5i − 3
0
1
0
from which we deduce that (i, 1, 0)T is an eigenvector. Let u3 = √12 (i, 1, 0)T .
If
i −i
1
P = √ 1 1
2 0 0
0
√0
2
and
3 0
D = 0 1
0 0
0
0 ,
5i
then P is a unitary matrix and P ∗ AP = D.
The matrix A is not Hermitian. Since it has a complex number on the diagonal,
A∗ ̸= A. However, it is normal since it can be unitarily diagonalised. (This can
also be shown directly by verifying that AA∗ = A∗ A.)
Problem 13.15 Suppose A is skew-Hermitian, so A∗ = −A. Let λ ̸= 0 be an
eigenvalue of A with corresponding unit eigenvector v.
CHAPTER 13.
152
To prove (1), consider
v∗ Av = v∗ (λv) = λ(v∗ v) = λ,
so that
(v∗ Av)∗ = λ∗ = λ.
On the other hand, since A is skew-Hermitian,
(v∗ Av)∗ = v∗ A∗ v = v∗ (−A)v = −v∗ (Av) = −v∗ (λv) = −λ(v∗ v) = −λ.
Comparing these, we have λ = −λ. If λ = a + ib, this implies that a = 0 (since
a − ib = −a − ib), so λ = ib is pure imaginary.
(2) If A is skew-Hermitian, the proof that eigenvectors corresponding to distinct
eigenvalues are mutually orthogonal is the same as for Hermitian matrices.
Suppose that λ and µ are any two different eigenvalues of A and that x, y are
corresponding eigenvectors, so that Ax = λx and Ay = µy. Then
y∗ Ax = y∗ λx = ⟨λx, y⟩ = λ⟨x, y⟩
On the other hand, since A = −A∗ .
y∗ Ax = −y∗ A∗ x = −(Ay)∗ x = −(µy)∗ x = −µy∗ x = −µ⟨x, y⟩.
But we have just shown that the eigenvalues of a Hermitian matrix are pure
imaginary, so µ = −µ, and we can conclude from these two expressions for
y∗ Ax that λ⟨x, y⟩ = µ⟨x, y⟩, or
(λ − µ)⟨x, y⟩ = 0.
Since λ ̸= µ, we deduce that ⟨x, y⟩ = 0. That is, x and y are orthogonal.
(3) To show that skew-Hermitian matrices are normal, we need to verify that if
A = −A∗ , then AA∗ = A∗ A. We have
AA∗ = A(−A) = −A2
and
A∗ A = (−A)A = −A2 ,
so that A is indeed normal.
Problem 13.16 It is straightforward to find A∗ and −A for the matrix
i 1 0
A = −1 0 −1
0 1 i
153
and check that they are equal.
To find the eigenvalues, find the roots of the characteristic equation. Expanding
by the first row, you should find that
|A − λI| =
i−λ 1
−1 −λ
0
1
0
−1 = (i − λ)(λ2 − λi + 2).
i−λ
Using the quadratic formula (or factorising the quadratic), the eigenvalues are
λ1 = i, λ2 = −i, λ3 = 2i. Note that these are all pure imaginary. For each
eigenvalue, find a corresponding eigenvector by solving (A − λI)x = 0. (The
details are omitted.) An orthonormal set of eigenvectors corresponding to
λ1 , λ2 , λ3 is given by
−1
1
1
1
1
1
, x2 = √
x1 = √
0
−2i , x3 = √
i .
2
6
3 1
1
1
You should check that these eigenvectors are mutually orthogonal unit vectors,
and that they are eigenvectors (with Axi = λi xi ).
Then the spectral decomposition of A is
i 1 0
A = −1 0 −1 = λ1 x1 x∗1 + λ2 x2 x∗2 + λ3 x3 x∗3
0 1 i
−1
1
1
i
2i
i
= 0 ( −1 0 1 ) − −2i ( 1 2i 1 ) + i ( 1 −i
2
6
3
1
1
1
1 0 −1
1
2i
1
1 −i 1
i
i
2i
= 0 0 0 − −2i 4 −2i + i 1 i .
2
6
3
−1 0 1
1
2i
1
1 −i 1
(You should check this by adding the matrices to obtain A.)
1)
CHAPTER 13.
154
Since A∗ = −A, we have
−i −1 0
A∗ = 1
0
1
0 −1 −i
1 0 −1
1
2i
i
i
= − 0 0 0 + −2i 4
2
6
−1 0 1
1
2i
1
1 −i 1
2i
−2i − i 1 i .
3
1
1 −i 1
Problem 13.17 (Note: there was a typographical error in the text. Instead of
Ei = xx∗ , it should have read Ei = xi x∗i .) If P is the matrix whose columns are
an orthonormal basis of C consisting of eigenvectors x1 , x2 , . . . , xn of A, we
already know that
A = λ1 E1 + λ2 E2 + · · · + λn En
where Ei = xi x∗i .
If x is an eigenvector of the matrix A corresponding to the eigenvalue λ, then it is
also an eigenvector of the matrix Ak corresponding to the eigenvalue λk , since if
Ax = λx, then
Ak x = Ak−1 Ax = Ak−1 λx = λAk−1 x = λAk−2 Ax = λ2 Ak−2 x = · · · = λk x.
Therefore the same reasoning applies (as for the spectral decomposition of A)
and we have
Ak = λk1 E1 + λk2 E2 + · · · + λkn En .
An alternative approach (which to be done properly, requires proof by induction)
uses the fact that Ei Ej = 0 if i ̸= j and Ei2 = Ei . We can argue, on that basis,
that
A2 = (λ1 E1 + λ2 E2 + · · · + λn En )(λ1 E1 + λ2 E2 + · · · + λn En )
= λ21 E12 + λ22 E22 + · · · + λ2n En2
= λ21 E1 + λ22 E2 + · · · + λ2n En ,
and so on.
On the other hand, since A = P DP ∗ where P is unitary, we have Ak = P Dk P ∗ .
So these expressions for Ak are equal; each one is equal to
Ak = λk1 x1 x∗1 + λk2 x2 x∗2 + · · · + λkn xn x∗n .
155
Looking at the solution of Problem 13.10, the eigenvalues of A are λ1 = 5,
λ2 = 1 and λ3 = 4, and the spectral decomposition of A is given by
A = λ1 u1 u∗1 + λ2 u2 u∗2 + λ3 u3 u∗3
1+i
1+i
0
4
1
= 5 0 ( 0 0 1 ) + −1 ( 1 − i −1 0 ) + 2 ( 1 − i 2 0 )
3
6
0
0
1
( )
0 0 0
2
−1 − i 0
2
2 + 2i 0
1
1
= 5 0 0 0 + −1 + i
1
0 + 4
2 − 2i
4
0
3
6
0 0 1
0
0
0
0
0
0
= 5E1 + E2 + 4E3 ,
which can be easily checked. Then A3 is given by
A3 = 53 E1 + E2 + 43 E3
2
0 0 0
1
= 125 0 0 0 + −1 + i
3
0
0 0 1
so
22
3
A = 21(1 − i)
0
−1 − i 0
2
64
1
0 +
2 − 2i
6
0
0
0
21(1 + i) 0
43
0 .
0
125
2 + 2i 0
4
0,
0
0
