“JUST THE MATHS”
UNIT NUMBER
9.9
MATRICES 9
(Modal & spectral matrices)
by
A.J.Hobson
9.9.1
9.9.2
9.9.3
9.9.4
Assumptions and definitions
Diagonalisation of a matrix
Exercises
Answers to exercises
UNIT 9.9 - MATRICES 9
MODAL AND SPECTRAL MATRICES
9.9.1 ASSUMPTIONS AND DEFINITIONS
For convenience, we shall make, here, the following assumptions:
(a) The n eigenvalues, λ1 , λ2 , λ3 , . . . , λn , of an n × n matrix, A, are arranged in order of
decreasing value.
(b) Corresponding to λ1 , λ2 , λ3 , . . . , λn , respectively, A possesses a full set of eigenvectors
X1 , X2 , X3 , . . . , Xn , which are linearly independent.
If two eigenvalues coincide, the order of writing down the corresponding pair of eigenvectors
will be immaterial.
DEFINITION 1
The square matrix obtained by using as its columns any set of linearly independent eigenvectors of a matrix A is called a “modal matrix” of A, and may be denoted by M.
Notes:
(i) There are infinitely many modal matrices for a given matrix, A, since any multiple of an
eigenvector is also an eigenvector.
(ii) It is sometimes convenient to use a set of normalised eigenvectors.
When using normalised eigenvectors, the modal matrix may be denoted by N and, for an n×n
matrix, A, there are 2n possibilities for N since each of the n columns has two possibilities.
DEFINITION 2
If λ1 , λ2 , λ3 , . . . , λn are the eigenvalues of an n × n matrix, A, then the diagonal matrix,
λ1
 0


 0

 .
0

0
λ2
0
.
0
0
0
λ3
.
0
.
.
.
.
.
. 0
. 0 


. 0 

. . 
. λn

is called the “spectral matrix” of A, and may be denoted by S.
1
EXAMPLE
For the matrix,


1 1 −2
A=
1 
 −1 2
,
0 1 −1
determine a modal matrix, a modal matrix of normalised eigenvectors and the spectral
matrix.
Solution
The characteristic equation is
1 − λ
−1
0
1
−2 2−λ
1 = 0,
1
−1 − λ which may be shown to give
−(1 + λ)(1 − λ)(2 − λ) = 0.
Hence, the eigenvalues are λ1 = 2, λ2 = 1 and λ3 = −1 in order of decreasing value.
Case 1. λ = 2
We solve the simultaneous equations
−x + y − 2z = 0,
−x + 0y + z = 0,
0x + y − 3z = 0,
which give x : y : z = 1 : 3 : 1
2
Case 2. λ = 1
We solve the simultaneous equations
0x + y − 2z = 0,
−x + y + z = 0,
0x + y − 2z = 0,
which give x : y : z = 3 : 2 : 1
Case 3. λ = −1
We solve the simultaneous equations
2x + y − 2z = 0,
−x + 3y + z = 0,
0x + y + 0z = 0,
which give x : y : z = 1 : 0 : 1
A modal matrix for A may therefore be given by


1 3 1


M = 3 2 0.
1 1 1
A modal matrix of normalised eigenvectors may be given by


N=

√1
11
√3
11
√1
11
√3
14
√2
14
√1
14
3
√1
2


0 
.
√1
2
The spectral matrix is given by


2 0 0


S = 0 1 0 .
0 0 −1
9.9.2 DIAGONALISATION OF A MATRIX
Since the eigenvalues of a diagonal matrix are equal to its diagonal elements, it is clear that
a matrix, A, and its spectral matrix, S, have the same eigenvalues.
From the Theorem in Unit 9.8, therefore, it seems reasonable that A and S could be similar
matrices; and this is the content of the following result which will be illustrated rather than
proven.
THEOREM
The matrix, A, is similar to its spectral matrix, S, the similarity transformation being
M−1 AM = S,
where M is a modal matrix for A.
ILLUSTRATION:
Suppose that X1 , X2 and X3 are linearly independent eigenvectors of a 3 × 3 matrix, A,
corresponding to eigenvalues λ1 , λ2 and λ3 , respectively.
Then,
AX1 = λ1 X1 , AX2 = λ2 X2 , and AX3 = λ3 X3 .
Also,

x1
M=
 y1
z1
x2
y2
z2

x3
y3 
.
z3
If M is premultiplied by A, we obtain a 3×3 matrix whose columns are AX1 , AX2 , and AX3 .
4
That is,
AM = [ AX1
AX2
AX3 ] = [ λ1 X1
λ 2 X2
λ 3 X3 ]
or

x1

AM =  y1
z1
 
x2
y2
z2
x3
λ1
 
y3  .  0
z3
0
0
λ2
0

0

0  = MS.
λ3
We conclude that
M−1 AM = S.
Notes:
(i) M−1 exists only because X1 , X2 and X3 are linearly independent.
(ii) The similarity transformation in the above theorem reduces the matrix, A, to “diagonal
form” or “canonical form” and the process is often referred to as the “diagonalisation”
of the matrix, A.
EXAMPLE
Verify the above Theorem for the matrix


1 1 −2


A =  −1 2 1  .
0 1 −1
Solution
From an earlier example, a modal matrix for A may be given by


1 3 1


M = 3 2 0.
1 1 1
5
The spectral matrix is given by


2 0 0


S = 0 1 0 .
0 0 −1
It may be shown that

M−1

2
1  −2 2

=  3
0 −3 
6 −1 −2 7
and, hence,

 
 

2
1 1 −2
1 3 1
1  −2 2
 
 

−1
0 −3  .  −1 2 1  .  3 2 0 
M AM =  3
6 −1 −2 7
0 1 −1
1 1 1

 

2
2 3 −1
1 −2 2

= 
0 −3 
0 
 3
.6 2

6 −1 −2 7
2 1 −1


2 0 0


=  0 1 0  = S.
0 0 −1
6
9.9.3 EXERCISES
1. Determine a modal matrix, M, of linearly independent eigenvectors for the matrix


−1
6
−12


A =  0 −13 30  .
0
−9
20
Verify that M−1 AM = S, where S is the spectral matrix of A.
2. Determine a modal matrix, M, of linearly independent eigenvectors for the matrix


1 2 2


B =  0 2 1.
−1 1 2
Verify that M−1 AM = S, where S is the spectral matrix of B.
3. Determine a modal matrix, N, of linearly independent normalised eigenvectors for the
matrix


2 −2 0


C = 0 4 0.
2 5 1
Verify that N−1 AN = S, where S is the spectral matrix of C.
4. Show that the following matrices are not similar to a diagonal matrix:


1 2 2
(a) 
2 1
 0
,
−1 2 2
7


0 1 0
(b) 
0 1
0
.
1 −3 3
9.9.4 ANSWERS TO EXERCISES
1. The eigenvalues are 5,2 and −1, which gives


−1 0 1


M =  5 2 0.
3 1 0
2. The eigenvalues are 2, 1 and −1, which gives


1 3 1


M = 3 2 0.
1 1 1
3. The eigenvalues are 4, 2 and 1, which gives

− √13
√1
5

√1
3
√1
3
√2
5
N=

0

0

0
.
1
4. (a) The eigenvalues are 2 (repeated) and 1 but there are only two linearly independent
eigenvectors, namely
 
2
 
1
0

and

1


 1 .
−1
(b) There is only one eigenvalue, 1 (repeated), and only one linearly independent
eigenvector, namely
 
1
 
1.
1
8