Mathematical Methods for Physics and Engineering - Matematica.NET

8.17 QUADRATIC AND HERMITIAN FORMSi.e. Q is unchanged by considering only the symmetric part of M. Hence, with noloss of generality, we may assume A = A T in (8.106).From its definition (8.105), Q is clearly a basis- (i.e. coordinate-) independentquantity. Let us therefore consider a new basis related to the old one by anorthogonal transformation matrix S, the components in the two bases of anyvector x being related (as in (8.91)) by x = Sx ′ or, equivalently, by x ′ = S −1 x =S T x. We then haveQ = x T Ax =(x ′ ) T S T ASx ′ =(x ′ ) T A ′ x ′ ,where (as expected) the matrix describing the linear operator A in the newbasis is given by A ′ = S T AS (since S T = S −1 ). But, from the last section, if wechoose as S the matrix whose columns are the normalised eigenvectors of A thenA ′ = S T AS is diagonal with the eigenvalues of A as the diagonal elements. (SinceA is symmetric, its normalised eigenvectors are orthogonal, or can be made so,and hence S is orthogonal with S −1 = S T .)In the new basisQ = x T Ax =(x ′ ) T Λx ′ = λ 1 x ′ 12 + λ 2 x ′ 22 + ···+ λ N x ′ N2 , (8.110)where Λ = diag(λ 1 ,λ 2 ,...,λ N )andtheλ i are the eigenvalues of A. It should benoted that Q contains no cross-terms of the form x ′ 1 x′ 2 .◮Find an orthogonal transformation that takes the quadratic form (8.107) into the formλ 1 x ′ 12 + λ 2 x ′ 22 + λ 3 x ′ 32 .The required transformation matrix S has the normalised eigenvectors of A as its columns.We have already found these in section 8.14, and so we can write immediately⎛ √ √ ⎞3S = √ 1 √ √ 2 1⎝ 3 − 2 −1 ⎠6 √,0 2 −2which is easily verified as being orthogonal. Since the eigenvalues of A are λ =2,3,and−6, the general result already proved shows that the transformation x = Sx ′ will carry(8.107) into the form 2x ′ 12 +3x ′ 22 − 6x ′ 32 . This may be verified most easily by writing outthe inverse transformation x ′ = S −1 x = S T x and substituting. The inverse equations arex ′ 1 =(x 1 + x 2 )/ √ 2,x ′ 2 =(x 1 − x 2 + x 3 )/ √ 3, (8.111)x ′ 3 =(x 1 − x 2 − 2x 3 )/ √ 6.If these are substituted into the form Q =2x ′ 12 +3x ′ 22 − 6x ′ 32 then the original expression(8.107) is recovered. ◭In the definition of Q it was assumed that the components x 1 , x 2 , x 3 and thematrix A were real. It is clear that in this case the quadratic form Q ≡ x T Ax is real289