Mathematical Methods for Physics and Engineering : A Comprehensive Guide

8.17 QUADRATIC AND HERMITIAN FORMS

as the necessary condition thatxmust satisfy. If (8.114) is satisfied for some

eigenvectorxthen the value ofQ(x) is given by

Q=xTAx=xTλx=λ. (8.115)

However, ifxandyare eigenvectors corresponding to different eigenvalues then

they are (or can be chosen to be) orthogonal. Consequently the expressionyTAx

is necessarily zero, since

yTAx=yTλx=λyTx=0. (8.116)

Summarising, those column matrices x of unit magnitude that make the

quadratic formQstationary are eigenvectors of the matrixA, and the stationary

value ofQis then equal to the corresponding eigenvalue. It is straightforward

to see from the proof of (8.114) that, conversely, any eigenvector ofAmakesQ

stationary.

Instead of maximising or minimisingQ=xTAxsubject to the constraint

xTx= 1, an equivalent procedure is to extremise the function

λ(x)=

xTAx

xTx

.

Show that ifλ(x)is stationary thenxis an eigenvector ofAandλ(x)is equal to the

corresponding eigenvalue.

We require ∆λ(x) = 0 with respect to small variations inx.Now

∆λ=

1

(xTx)^2

[

(xTx)

(

∆xTAx+xTA∆x

)

−xTAx

(

∆xTx+xT∆x

)]

=

2∆xTAx

xTx

− 2

(

xTAx

xTx

)

∆xTx

xTx

,

sincexTA∆x=(∆xT)AxandxT∆x=(∆xT)x. Thus

∆λ=

2

xTx

∆xT[Ax−λ(x)x].

Hence, if ∆λ=0thenAx=λ(x)x,i.e.xis an eigenvector ofAwith eigenvalueλ(x).

Thus the eigenvalues of a symmetric matrixAare the values of the function

λ(x)=

xTAx

xTx

at its stationary points. The eigenvectors ofAlie along those directions in space

for which the quadratic formQ=xTAxhas stationary values, given a fixed

magnitude for the vectorx. Similar results hold for Hermitian matrices.