MATRICES AND VECTOR SPACES
Construct an orthonormal set of eigenvectors for the matrixA=
103
0 − 20
301
.
We first determine the eigenvalues using|A−λI|=0:
0=
∣
∣∣
∣
∣∣
1 −λ 03
0 − 2 −λ 0
301 −λ∣
∣∣
∣
∣∣=−(1−λ)(^2) (2 +λ) + 3(3)(2 +λ)
=(4−λ)(λ+2)^2.
Thusλ 1 =4,λ 2 =−2=λ 3. The eigenvectorx^1 =(x 1 x 2 x 3 )Tis found from
103
0 − 20
301
x 1
x 2
x 3
=4
x 1
x 2
x 3
⇒ x^1 =√^1
2
1
0
1
.
A general column vector that is orthogonal tox^1 is
x=(ab−a)T, (8.89)and it is easily shown that
Ax=
103
0 − 20
301
a
b
−a
=− 2
a
b
−a
=− 2 x.Thusxis a eigenvector ofAwith associated eigenvalue−2. It is clear, however, that there
is an infinite set of eigenvectorsxall possessing the required property; the geometrical
analogue is that there are an infinite number of corresponding vectorsxlying in the
plane that hasx^1 as its normal. We do require that the two remaining eigenvectors are
orthogonal to one another, but this still leaves an infinite number of possibilities. Forx^2 ,
therefore, let us choose a simple form of (8.89), suitably normalised, say,
x^2 =( 010 )T.The third eigenvector is then specified (to within an arbitrary multiplicative constant)
by the requirement that it must be orthogonal tox^1 andx^2 ; thusx^3 may be found by
evaluating the vector product ofx^1 andx^2 and normalising the result. This gives
x^3 =1
√
2
(− 101 )T,
to complete the construction of an orthonormal set of eigenvectors.
8.15 Change of basis and similarity transformationsThroughout this chapter we have considered the vectorxas a geometrical quantity
that is independent of any basis (or coordinate system). If we introduce a basis
ei,i=1, 2 ,...,N, into ourN-dimensional vector space then we may write
x=x 1 e 1 +x 2 e 2 +···+xNeN,