FACULTY OF MATHEMATICAL STUDIES
MATHEMATICS FOR PART I ENGINEERING
Lectures
MODULE 18
MATRICES III
1. Rank of a matrix
2. General systems of linear equations
3. Eigenvalues and eigenvectors
1. Rank of a matrix
Sets of n simultaneous linear equations in n unknowns (in the form AX = b) were considered in module
17. When the determinant of A is non-zero (i.e. |A| =
6 0) it proved easy to obtain solutions. However,
when |A| = 0 it is much harder, and to clarify the position it is helpful to introduce the concept of the
rank of a matrix.
Consider the system AX = b . To solve this system using the elimination method discussed in module 17
you reduce the matrix A to upper triangular form with non-zero elements along the principal diagonal.
This procedure is possible when |A| =
6 0 . However, the rearrangement fails when |A| = 0 , since in this
case the matrix can only be reduced to echelon form. An example of a matrix in echelon form is shown
below, with the dots denoting any non-zero quantities whereas the letters denote quantities which can be
zero or non-zero:


· a b c d
0 0 · e f 

.
0 0 0 0 ·
0 0 0 0 0
Observe that in this echelon form the first non-zero elements in each row do not all lie on the principal
diagonal, as they do when the matrix is in upper triangular form.
Def.
The number of non-zero rows in the echelon form of a matrix A equals the rank of A .
In the above example the rank of the matrix is 3. It is important to point out that rank is defined for
matrices of all orders, not just square matrices.
The rank of a matrix gives the number of essentially different rows in the matrix (i.e. rows which cannot
be written as linear combinations of some or all of the other rows). Understanding this point is important
in understanding the different types of solution that are possible for general sets of equations of the form
AX = b .
[The usual mathematical definition of rank is that it is the number of rows in the largest square sub-matrix
with non-zero determinant. A square sub-matrix is found from A by deleting a number of rows and columns
to obtain a square matrix. It is possible to show that the two definitions of rank are identical. Calculating
the rank using sub-matrices can be tricky when the matrix is large, and in practice it is usually better to
adopt the echelon approach.]
For a system AX = b , define the augmented matrix (A : b) as the matrix A with the column vector
b added to it as an extra column. With this definition we can state the following result.
Result
If rank A 6= rank (A : b) then AX = b has NO solution (equations are inconsistent);
if rank A = rank (A : b) then AX = b has a solution (equations are consistent) and this solution has s
free parameters (where s = number of unknowns − rank A).
1

Ex 1.
1
0

Find the rank of the matrix A =  1

2
2
0
1
1
3
2

−1
1 −1
1 −1
1

0
0
0 .

1 −1
1
0
0
0
We now use the elimination process to reduce the matrix A to echelon form, by considering each column
in turn. Leaving the first row unaltered we first ensure the remaining non-zero elements in the first column
become zero; then leaving the first two rows unaltered we make zero any non-zero elements in the bottom
part of the second column, continuing this process column by column until the matrix is in echelon form.
With this procedure

A
→
1
0

0

0
0

→
1
0

0

0
0
0
1
1
3
2
−1
1
1
3
2

1 −1
−1
1

−1
1,

−3
3
−2
2
row 3 − row 1
row 4 − 2 × row 1
row 5 − 2 × row 1
0
1
0
0
0

−1
1 −1
1 −1
1

0
0
0,

0
0
0
0
0
0
row 3 − row 2
row 4 − 3 × row 2
row 5 − 2 × row 2
The matrix is now in echelon form and the rank is the number of non-zero rows. Hence, it follows that
rank A = 2.
2. General systems of equations
In real situations the systems will contain a large number of equations, but to illustrate the ideas let us
consider very small systems.
(a) Under-specified sets of equations (number of equations less than number of unknowns)
 
x
1
1 1 1  
.
y =
Case (i)
Solve
2
1 2 3
z
This set is under-specified because when the matrices are multiplied out we have two equations in three
unknowns. Forming the augmented matrix, and then reducing it to echelon form gives
1 1
1 2
1 1
3 2
→
1 1
0 1
1
2
1
1
,
row 2 − row 1 .
The rank of the augmented matrix, the number of non-zero rows in the full reduced matrix above, is clearly
2. Ignoring the final column gives the matrix A and the number of non-zero rows in the reduced form of
this is again 2. Since the ranks are equal, the set of equations is consistent.
The reduced set of equations can be read off from the reduced form of the augmented matrix, the first three
columns giving the coefficients of the unknowns on the LHS, whereas the final column provides the new
column vector on the RHS. Thus the reduced system is
x + y + z = 1,
y + 2z = 1.
2
Put z = C, a constant, then the second equation implies y = 1 − 2C . Substituting these quantities into
the first of the reduced equations then yields x = 1 − y − z = 1 − (1 − 2C) − C = C . Thus the solution is
  

x
C
 y  =  1 − 2C ,
z
C
for any constant C .
[Note that in the above example the number of unknowns in the original set of equations = 3, rank A = 2
and so the general result stated earlier implies there is one free parameter, namely the constant C in the
above solution.]
 
x
1 1 1  
1
Case (ii)
y =
.
Solve
2 2 2
1
z
Here again there are two equations in three unknowns. Reducing the augmented matrix gives
1 1
2 2
1
2
1
1
→
1
0
1 1
0 0
1
−1
,
row 2 − 2 × row 1 .
In this case the rank of the augmented matrix is 2, but the rank of the matrix A is 1. These values are
not the same and so the system does NOT have a solution – the set of equations is inconsistent (which is
obvious, of course, from the original equations).
(b) Over-specified sets of equations (more equations than unknowns)




−1
1 1 x
=  0 .
Solve  1 2 
Case (iii)
y
1
1 3
The system has three equations in two unknowns. Reduction of the augmented matrix leads to

1 1
1 2
1 3

−1
0
1

→
→
1
0
0

1
0
0

1 −1
1
1 ,
2
2

1 −1
1
1 ,
0
0
row 2 - row 1
row 3 - row 1
row 3 − 2 × row 2 .
Hence rank A = 2 and rank (A : b) = 2 , the same, and so the equations are consistent. The reduced
system is
x + y = −1,
y = 1,
0=
0.
Hence, it follows that y = 1 and x = −1 − y = −1 − 1 = −2 , and the solution is
−2
x
.
=
1
y
[Number of original unknowns =2, rank A = 2 , so there are 2 − 2 = 0 free parameters in the above solution.]
In all cases the rank of the augmented matrix gives the number of independent equations in the given system,
i.e. equations which cannot be obtained as combinations of other equations in the system.
3



0
1 1 x
=  1 .
Case (iv)
Solve  1 2 
y
−2
1 3
Again there are three equations in two unknowns.
to



1
1 1
0
1 2
1 → 0
0
1 3 −2

1
→ 0
0

In this case manipulation of the augmented matrix leads

1
0
row 2 − row 1
1
1 ,
row 3 − row 1
2 −2

1
0
1
1 ,
row 3 − 2 × row 2 .
0 −4
Thus rank A = 2 and rank (A : b) = 3 , which are different. The equations are inconsistent with no
solution.
For the above four systems of equations the types of solution are easily spotted. However, the situation is
much less obvious if you consider larger systems (say 10 equations in 9 unknowns). The procedure discussed
above provides a framework for determining the appropriate solution for systems of all sizes.
3. Eigenvalues and eigenvectors
The theory of eigenvalues is of importance in many physical problems, e.g. considering the stability of
numerical methods for solving differential equations and determining the normal modes of oscillation of
vibrating systems.
(a) 2 × 2 matrix
Let us illustrate the method with a 2 × 2 matrix A. Suppose
x1
a11 a12
,
X=
,
A=
a21 a22
x2
then consider the eigenvalue equation
AX = λX,
(1)
where λ is a scalar (a number). Equation (1) can be written
AX − λX = 0,
→
AX − λIX = (A − λI)X = 0,
(2)
where I is the unit 2 × 2 matrix. Written in full (2) is a set of two simultaneous linear equations in two
unknowns:
a12 x2 = 0,
(a11 − λ) x1 +
(3)
a21 x1 + (a22 − λ) x2 = 0.
Obviously the two equations have the trivial solution x1 = x2 = 0 (i.e. X = 0) but this solution is usually
of no physical interest. However, for particular values of λ non-zero solutions for x1 and x2 may exist and
these often do have physical significance. Equations (3) can be written BX = 0, where B = A−λI, and you
know from module 17 that for this set of equations to have a non-zero solution it is necessary that |B| = 0,
i.e. |A − λI| = 0.
For the 2 × 2 matrix A , condition (4) becomes
a11 − λ
a21
a12 = 0,
a22 − λ i.e. (a11 − λ)(a22 − λ) − a12 a21 = 0,
4
(4)
→
λ2 − (a11 + a22 )λ + a11 a22 − a12 a21 = 0.
(5)
This is a quadratic equation in λ, known as the characteristic equation of A, and the left-hand side
is called the characteristic polynomial of A. The two solutions of equation (5), λ1 , λ2 say (normally
distinct), are called the eigenvalues of A. (In some texts the word eigenvalue is replaced by characteristic
value or latent root). When the solution λ = λ1 is substituted back into the system (3), the resulting
equations can be solved to produce non-zero solutions for x1 and x2 , called “the eigenvector corresponding
to the eigenvalue λ1 ”. In a similar way you obtain the eigenvector corresponding to the eigenvalue λ2 .
Let us now illustrate the above procedure for a particular 2 × 2 matrix.
1 2
.
Find the eigenvalues and eigenvectors of the matrix A =
Ex 2.
0 3
We need to solve the matrix equation AX =λX, or (A − λI)X = 0. Non-zero solutions for λ exist only if
|A − λI| = 0,
1− λ
2 = 0,
i.e. 0
3 − λ
(1 − λ)(3 − λ) − 0(2) = 0,
(1 − λ)(3 − λ) = 0,
yielding λ = 1, 3
(these are the eigenvalues).
We now calculate in turn the eigenvector corresponding to each eigenvalue. For λ = 1 it is necessary to solve
(A−1I)X = 0,
x1
1−1
2
0
,
=
0
3−1
0
x2
0 2
x1
0
=
,
0 2
x2
0
which gives the two equations 2x2 = 0 and 2x2 = 0. These are obviously identical and both are satisfied
when x2 = 0. There
restriction
on x1 , however, which can have any value, = C say. Hence the
isno 1
C
x1
=C
=
, for any constant C.
eigenvector is
0
0
x2
For the other eigenvalue λ = 3, a similar procedure is adopted. The system (A−3 I)X = 0 corresponds to
1−3
2
x1
0
=
,
x2
0
3−3
0
x1
−2 2
0
,
=
0 0
0
x2
leading to −2x1 + 2x2 = 0,
0. Here
choosex2= D, an arbitrary constant, then it follows that x1 = D.
0=
D
1
x1
=
=D
, for any constant D.
Hence the eigenvector is
x2
D
1
Two comments should be made about the above solution. Firstly, eigenvectors always contain at least one
free parameter, similar to C and D in the above example. If your answer does not possess this feature you
have made a mistake.
Secondly, you can always check that the answers satisfy the eigenvalue equation, AX = λX. In the above
example, for instance, the eigenvector corresponding to the eigenvalue 3 satisfies
1 2
D
D + 2D
3D
D
AX =
=
=
=3
= 3X = λX.
0 3
D
0 + 3D
3D
D
Similarly, for the other eigenvector you obtain
C
C
C+0
C
1 2
.
=1
=
=
0
0
0+0
0
0 3
5
(b) n × n matrix The theory presented above for a 2 × 2 matrix can easily be extended to a general n × n
matrix. Let A be an n × n matrix and X be an n × 1 column matrix defined by


a1n
a2n 
.. 
. 
a11
 a21
A=
 ...
a12
a22
..
.
...
...
..
.
an1
an2
. . . ann

and

x1
 x2 

X=
 ...  .
xn
Then the equation AX = λX, where λ is a number, becomes (A − λI) X = 0, where I is now the unit
n × n matrix. Written out in full we have a set of n simultaneous homogeneous equations in the n variables
x1 , x2 , . . . , xn :
a12 x2 + . . . +
a1n xn = 0
(a11 − λ)x1 +
a2n xn = 0
a21 x1 + (a22 − λ)x2 + . . . +
(6)
..
..
..
..
.
.
.
.
an1 x1
+
an2 x2
+
...
+
(ann − λ)xn
=
0
The above system always has the trivial solution x1 = x2 = . . . = 0 (i.e. X = 0) but non-zero solutions can
exist for values of λ which satisfy:
a12
...
a1n a11 − λ
a22 − λ . . .
a2n a21
= 0.
|A − λI| = ..
..
..
..
.
.
.
.
an1
an2
. . . ann − λ
(7)
Condition (7) again leads to the characteristic equation of A, which is now a polynomial equation of
degree n in λ. The n roots λi (i = 1, 2 . . . n) of (7) (which are not necessarily all distinct) are called the
eigenvalues of A. If a particular eigenvalue λi is substituted back into (6) the resulting equations can be
solved for x1 , x2 , . . . , xn to give the eigenvector (always non-zero) corresponding to the eigenvalue λi .
Let us work through an example for a 3 × 3 matrix.


0
2
3
Ex 3. Find the eigenvalues and eigenvectors of the matrix A =  1 −2 −1  .
2
1
1
Need to solve |A − λI| = 0 ,
i.e.
−λ
2
3 1 −2 − λ
−1 = 0 .
2
1
1 − λ
We know the characteristic equation here will be cubic (since A is a 3 × 3 matrix) and there are two ways
to proceed. The determinant can be expanded directly but in order to find the eigenvalues it will then
be necessary to determine the roots of the third order polynomial. This is usually not too difficult, but it
usually involves spotting factors and can be tricky. The alternative approach is to manipulate the elements
in the determinant before the evaluation in order to simplify the calculations and pull out common factors
(involving λ ). Both methods will be given below.
Direct way Expanding the determinant by the first row
−λ {(−2 − λ)(1 − λ) − (−1)1} − 2 {1(1 − λ) − (−1)2} + 3 {1(1) − (−2 − λ)2} = 0,
−λ{λ2 + λ − 2 + 1} − 2{1 − λ + 2} + 3{1 + 4 + 2λ} = 0,
−λ(λ2 + λ − 1) − 2(3 − λ) + 3(2λ + 5) = 0,
i.e.
− λ3 − λ2 + λ − 6 + 2λ + 6λ + 15 = 0,
λ3 + λ2 − 9λ − 9 = 0.
6
This cubic must now be factorised, so it is usually necessary to spot a value of λ which satisfies the equation,
say λ = a , then use the remainder theorem which states that λ − a is then a factor of the cubic. Here you
could spot that λ = −1 satisfies the cubic, but it is easier to observe that the cubic can be rearranged to
display a common factor:
λ3 + λ2 − 9λ − 9 = λ2 (λ + 1) − 9(λ + 1) = (λ + 1)(λ2 − 9)
therefore
(λ + 1)(λ2 − 9) = 0
(λ + 1)(λ − 3)(λ + 3) = 0
i.e.
λ = −1, 3, −3 (three eigenvalues)
Alternative method We spot that adding together the first and third columns of |A − λI| produces a zero
element and two others with a common factor dependent on λ, so
3−λ
2
3 col.1 + col.3
−2 − λ
−1 ,
|A − λI| = 0
3−λ
1
1 − λ
0
1
1
2
3 −1 = (3 − λ) 0 −2 − λ
= (3 − λ) 0 −2 − λ
1
1
1
1
1 − λ
2 + λ −1 , row 1-row 3.
1 − λ
Notice that we have carefully recorded the row and column operations we have carried out–very useful if you
need to check your working subsequently!
Only one element now remains in the first column and a common factor has already been extracted, hence
|A − λI| = (3 − λ)(−1)3+1 1(−1) − (2 + λ)(−2 − λ) = (3 − λ) − 1 + (2 + λ)2
= (3 − λ)(−1 + 4 + 4λ + λ2 ) = (3 − λ)(3 + 4λ + λ2 )
= (3 − λ)(1 + λ)(3 + λ) ,
and the same eigenvalues are derived as using the direct method.
Whichever way you have found the eigenvalues, the eigenvectors must be calculated by solving the matrix
equation (A − λI)X = 0.
λ = −3 Need to solve (A − (−3)I)X = 0, which can be written (A + 3I)X = 0. The set of equations to
be solved, found by adding 3 to the diagonal elements, is
3x1 + 2x2 + 3x3 = 0,
x1 + x2 − x3 = 0,
2x1 + x2 + 4x3 = 0.
It is convenient to interchange the first and second equations giving
x1 + x2 − x3 = 0,
3x1 + 2x2 + 3x3 = 0,
2x1 + x2 + 4x3 = 0.
Then using (eqn.2 – 3 × eqn.1) and (eqn.3 – 2 × eqn.1) the system becomes
x1 + x2 − x3 = 0,
− x2 + 6x3 = 0,
− x2 + 6x3 = 0.
7
Clearly the second and third equations are identical, so the last one can be ignored. If we choose x3 = C,
where C is any non-zero constant, then the second equation implies x2 = 6C and, after substitution into the
first equation, x1 = −5C. Hence the eigenvector corresponding to λ = −3 is

 
−5C
x1
 x2  =  6C  .
C
x3

λ = −1
Solve (A − (−1)I)X = 0, or (A + I)X = 0, equivalent to
x1 + 2x2 + 3x3 = 0,
x1 − x2 − x3 = 0,
2x1 + x2 + 2x3 = 0.
Subtract the first equation from the second equation, and 2× eqn.1 from the third equation, to give
x1 + 2x2 + 3x3 = 0,
− 3x2 − 4x3 = 0,
− 3x2 − 4x3 = 0.
Again the last two equations are identical. On putting x3 = D we find that x2 = − 34 D, and then x1 = − 13 D,
so that the eigenvector corresponding to λ = −1 is

 

x1
− D/3
 x2  =  −4D/3  ,
x3
D
for any D.
λ = 3 In this case we must solve (A−3I)X = 0. For the two previous eigenvalues we found the eigenvectors
by manipulating equations, but we could easily work with a matrix representation. We shall illustrate the
matrix method for this third eigenvalue. Thus, we need to find the solution of
   
0
−3
2
3
x1
 1 −5 −1   x2  =  0  ,
x3
0
2
1 −2

and interchanging the first and second rows (equivalent to interchanging the first and second equations) gives
   
0
1 −5 −1
x1
 −3
2
3   x2  =  0  .
x3
0
2
1 −2

Simplifying using the elimination method, by calculating (row 2 + 3 × row 1) and (row 3 - 2 × row 1), leads
to

   
0
1 −5 −1
x1
 0 −13
0   x2  =  0  .
x3
0
0
11
0
This is equivalent to the reduced system
x1 − 5x2 − x3 = 0
− 13x2
+ 11x2
8
=0
= 0.
The last pair of equations imply x2 = 0, and hence using the first equation x1 = x3 . Choosing x3 = E we
obtain
   
E
x1
 x2  =  0  , for any E .
E
x3
It is important to note that there is more than one way in which to write the eigenvectors. For instance,
in finding the eigenvector corresponding to λ = −3 in the above Example we could have chosen x2 = F ,
in which case it follows that x3 = F/6 and hence x1 = −5F/6. In this case it is easily seen that the new
solution is identical to the previous one after a redefinition of the arbitrary constant (F = 6C).
Concluding comments Before finishing the module two further points are briefly mentioned.
(i) It can be shown that if A is symmetric, a situation which occurs in many applications, then the
eigenvalues are real and the eigenvectors corresponding to different eigenvalues are perpendicular.
(ii) In all the cases considered in this module the roots of the characteristic equation are real and distinct.
Clearly this is not always the case. When some, or all, of the roots are equal then the calculation of the
eigenvectors is more complicated. This more difficult case is not considered in this module.
rec/00lm3
9