8.13 EIGENVECTORS AND EIGENVALUES
writeA†xi=λ∗ixi. From (8.74) and (8.75) we have
(λi−λj)(xi)†xj= 0. (8.76)
Thus,ifλi=λjthe eigenvectorsxiandxjmust be orthogonal,i.e.(xi)†xj= 0.
It follows immediately from (8.76) that if allNeigenvalues of a normal matrix
Aare distinct then allNeigenvectors ofAare mutually orthogonal. If, however,
two or more eigenvalues are the same then further consideration is required. An
eigenvalue corresponding to two or more different eigenvectors (i.e. they are not
simply multiples of one another) is said to bedegenerate. Suppose thatλ 1 isk-fold
degenerate, i.e.
Axi=λ 1 xi fori=1, 2 ,...,k, (8.77)
but that it is different from any ofλk+1,λk+2, etc. Then any linear combination
of thesexiis also an eigenvector with eigenvalueλ 1 ,since,forz=
∑k
i=1cix
i,
Az≡A
∑k
i=1
cixi=
∑k
i=1
ciAxi=
∑k
i=1
ciλ 1 xi=λ 1 z. (8.78)
If thexidefined in (8.77) are not already mutually orthogonal then we can
construct new eigenvectorszithat are orthogonal by the following procedure:
z^1 =x^1 ,
z^2 =x^2 −
[
(zˆ
1
)†x^2
]
zˆ^1 ,
z^3 =x^3 −
[
(zˆ
2
)†x^3
]
zˆ^2 −
[
(ˆz
1
)†x^3
]
zˆ^1 ,
..
.
zk=xk−
[
(zˆk−^1 )†xk
]
zˆk−^1 −···−
[
(zˆ^1 )†xk
]
ˆz^1.
In this procedure, known asGram–Schmidt orthogonalisation, each new eigen-
vectorziis normalised to give the unit vectorzˆibefore proceeding to the construc-
tion of the next one (the normalisation is carried out by dividing each element of
the vectorziby [(zi)†zi]^1 /^2 ). Note that each factor in brackets (ˆzm)†xnis a scalar
product and thus only a number. It follows that, as shown in (8.78), each vector
ziso constructed is an eigenvector ofAwith eigenvalueλ 1 and will remain so
on normalisation. It is straightforward to check that, provided the previous new
eigenvectors have been normalised as prescribed, eachziis orthogonal to all its
predecessors. (In practice, however, the method is laborious and the example in
subsection 8.14.1 gives a less rigorous but considerably quicker way.)
Therefore, even ifAhas some degenerate eigenvalues we canby construction
obtain a set ofNmutually orthogonal eigenvectors. Moreover, it may be shown
(although the proof is beyond the scope of this book) that these eigenvectors
arecompletein that they form a basis for theN-dimensional vector space. As