Determination of eigenvalues and eigenvectors

MATRICES AND VECTOR SPACES
8.14 Determination of eigenvalues and eigenvectors
The next step is to show how the eigenvalues and eigenvectors of a given N × N
matrix A are found. To do this we refer to (8.68) and as in (8.69) rewrite it as
Ax − λIx = (A − λI)x = 0.
(8.85)
The slight rearrangement used here is to write x as Ix, where I is the unit matrix
of order N. The point of doing this is immediate since (8.85) now has the form
of a homogeneous set of simultaneous equations, the theory of which will be
developed in section 8.18. What will be proved there is that the equation Bx = 0
only has a non-trivial solution x if |B| = 0. Correspondingly, therefore, we must
have in the present case that
|A − λI| = 0,
(8.86)
if there are to be non-zero solutions x to (8.85).
Equation (8.86) is known as the characteristic equation for A and its LHS as
the characteristic or secular determinant of A. The equation is a polynomial of
degree N in the quantity λ. The N roots of this equation λi , i = 1, 2, . . . , N, give
the eigenvalues of A. Corresponding to each λi there will be a column vector xi ,
which is the ith eigenvector of A and can be found by using (8.68).
It will be observed that when (8.86) is written out as a polynomial equation in
λ, the coefficient of −λN−1 in the equation will be simply A11 + A22 + · · · + ANN
relative to the coefficient of λN . As discussed in section 8.8, the quantity N
i=1 Aii
is the trace of A and, from the ordinary theory of polynomial equations, will be
equal to the sum of the roots of (8.86):
N
λi = Tr A.
(8.87)
i=1
This can be used as one check that a computation of the eigenvalues λi has been
done correctly. Unless equation (8.87) is satisfied by a computed set of eigenvalues,
they have not been calculated correctly. However, that equation (8.87) is satisfied is
a necessary, but not sufficient, condition for a correct computation. An alternative
proof of (8.87) is given in section 8.16.
Find the eigenvalues and normalised eigenvectors of the real symmetric matrix


1
1
3
1
−3  .
A= 1
3 −3 −3
Using (8.86),
1−λ
1
3
1
1−λ
−3
280
3
−3
−3 − λ
= 0.
8.14 DETERMINATION OF EIGENVALUES AND EIGENVECTORS
Expanding out this determinant gives
(1 − λ) [(1 − λ)(−3 − λ) − (−3)(−3)] + 1 [(−3)(3) − 1(−3 − λ)]
+ 3 [1(−3) − (1 − λ)(3)] = 0,
which simplifies to give
(1 − λ)(λ2 + 2λ − 12) + (λ − 6) + 3(3λ − 6) = 0,
⇒
(λ − 2)(λ − 3)(λ + 6) = 0.
Hence the roots of the characteristic equation, which are the eigenvalues of A, are λ1 = 2,
λ2 = 3, λ3 = −6. We note that, as expected,
λ1 + λ2 + λ3 = −1 = 1 + 1 − 3 = A11 + A22 + A33 = Tr A.
For the first root, λ1 = 2, a suitable eigenvector x1 , with elements x1 , x2 , x3 , must satisfy
Ax1 = 2x1 or, equivalently,
x1 + x2 + 3x3 = 2x1 ,
x1 + x2 − 3x3 = 2x2 ,
3x1 − 3x2 − 3x3 = 2x3 .
(8.88)
These three equations are consistent (to ensure this was the purpose in finding the particular
values of λ) and yield x3 = 0, x1 = x2 = k, where k is any non-zero number. A suitable
eigenvector would thus be
x1 = (k k 0)T .
√
If we apply the normalisation condition, we require k 2 + k 2 + 02 = 1 or k = 1/ 2. Hence
T
1
1
1
√
x1 = √
0
= √ (1 1 0)T .
2
2
2
Repeating the last paragraph, but with the factor 2 on the RHS of (8.88) replaced
successively by λ2 = 3 and λ3 = −6, gives two further normalised eigenvectors
1
x2 = √ (1
3
− 1 1)T ,
1
x3 = √ (1
6
−1
− 2)T . In the above example, the three values of λ are all different and A is a
real symmetric matrix. Thus we expect, and it is easily checked, that the three
eigenvectors are mutually orthogonal, i.e.
1 T 2 1 T 3 2 T 3
x = x
x = x
x = 0.
x
It will be apparent also that, as expected, the normalisation of the eigenvectors
has no effect on their orthogonality.
8.14.1 Degenerate eigenvalues
We return now to the case of degenerate eigenvalues, i.e. those that have two or
more associated eigenvectors. We have shown already that it is always possible
to construct an orthogonal set of eigenvectors for a normal matrix, see subsection 8.13.1, and the following example illustrates one method for constructing
such a set.
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