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)(λ^2 +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 ofA,areλ 1 =2,
λ 2 =3,λ 3 =−6. We note that, as expected,
λ 1 +λ 2 +λ 3 =−1=1+1−3=A 11 +A 22 +A 33 =TrA.
For the first root,λ 1 = 2, a suitable eigenvectorx^1 , with elementsx 1 ,x 2 ,x 3 ,mustsatisfy
Ax^1 =2x^1 or, equivalently,
x 1 +x 2 +3x 3 =2x 1 ,
x 1 +x 2 − 3 x 3 =2x 2 , (8.88)
3 x 1 − 3 x 2 − 3 x 3 =2x 3.These three equations are consistent (to ensure this was the purpose in finding the particular
values ofλ)andyieldx 3 =0,x 1 =x 2 =k,wherekis any non-zero number. A suitable
eigenvector would thus be
x^1 =(kk 0 )T.If we apply the normalisation condition, we requirek^2 +k^2 +0^2 =1ork=1/
√
- Hence
x^1 =(
1
√
2
1
√
2
0
)T
=
1
√
2
( 110 )T.
Repeating the last paragraph, but with the factor 2 on the RHS of (8.88) replaced
successively byλ 2 =3andλ 3 =−6, gives two further normalised eigenvectors
x^2 =1
√
3
( 1 − 11 )T, x^3 =1
√
6
( 1 − 1 − 2 )T.
In the above example, the three values ofλare all different andAis areal symmetric matrix. Thus we expect, and it is easily checked, that the three
eigenvectors are mutually orthogonal, i.e.
(
x^1)T
x^2 =(
x^1)T
x^3 =(
x^2)T
x^3 =0.It will be apparent also that, as expected, the normalisation of the eigenvectors
has no effect on their orthogonality.
8.14.1 Degenerate eigenvaluesWe 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 subsec-
tion 8.13.1, and the following example illustrates one method for constructing
such a set.