Mathematical Methods for Physics and Engineering : A Comprehensive Guide

8.17 QUADRATIC AND HERMITIAN FORMS

i.e.Qis unchanged by considering only the symmetric part ofM. Hence, with no

loss of generality, we may assumeA=ATin (8.106).

From its definition (8.105),Qis clearly a basis- (i.e. coordinate-) independent

quantity. Let us therefore consider a new basis related to the old one by an

orthogonal transformation matrixS, the components in the two bases of any

vectorxbeing related (as in (8.91)) byx=Sx′or, equivalently, byx′=S−^1 x=

STx. We then have

Q=xTAx=(x′)TSTASx′=(x′)TA′x′,

where (as expected) the matrix describing the linear operatorA in the new

basis is given byA′=STAS(sinceST=S−^1 ). But, from the last section, if we

choose asSthe matrix whose columns are thenormalisedeigenvectors ofAthen

A′=STASis diagonal with the eigenvalues ofAas the diagonal elements. (Since

Ais symmetric, its normalised eigenvectors are orthogonal, or can be made so,

and henceSis orthogonal withS−^1 =ST.)

In the new basis

Q=xTAx=(x′)TΛx′=λ 1 x′ 1

2

+λ 2 x′ 2

2

+···+λNx′N

2

, (8.110)

where Λ = diag(λ 1 ,λ 2 ,...,λN)andtheλiare the eigenvalues ofA. It should be

noted thatQcontains no cross-terms of the formx′ 1 x′ 2.

Find an orthogonal transformation that takes the quadratic form (8.107) into the form

λ 1 x′ 12 +λ 2 x′ 22 +λ 3 x′ 32.

The required transformation matrixShas thenormalisedeigenvectors ofAas its columns.
We have already found these in section 8.14, and so we can write immediately

S=

1

√

6





√

3

√

√^21

3 −

√

2 − 1

0

√

2 − 2



,

which is easily verified as being orthogonal. Since the eigenvalues ofAareλ=2,3,and
−6, the general result already proved shows that the transformationx=Sx′will carry
(8.107) into the form 2x′ 12 +3x′ 22 − 6 x′ 32 .This may be verified most easily by writing out
the inverse transformationx′=S−^1 x=STxand substituting. The inverse equations are

x′ 1 =(x 1 +x 2 )/

√

2 ,

x′ 2 =(x 1 −x 2 +x 3 )/

√

3 , (8.111)

x′ 3 =(x 1 −x 2 − 2 x 3 )/

√

6.

If these are substituted into the formQ=2x′ 12 +3x′ 22 − 6 x′ 32 then the original expression
(8.107) is recovered.

In the definition ofQit was assumed that the componentsx 1 ,x 2 ,x 3 and the

matrixAwere real. It is clear that in this case the quadratic formQ≡xTAxis real