Mathematical Methods for Physics and Engineering - Matematica.NET

8.17 QUADRATIC AND HERMITIAN FORMSas the necessary condition that x must satisfy. If (8.114) is satisfied for someeigenvector x then the value of Q(x) is given byQ = x T Ax = x T λx = λ. (8.115)However, if x and y are eigenvectors corresponding to different eigenvalues thenthey are (or can be chosen to be) orthogonal. Consequently the expression y T Axis necessarily zero, sincey T Ax = y T λx = λy T x =0. (8.116)Summarising, those column matrices x of unit magnitude that make thequadratic form Q stationary are eigenvectors of the matrix A, and the stationaryvalue of Q is then equal to the corresponding eigenvalue. It is straightforwardto see from the proof of (8.114) that, conversely, any eigenvector of A makes Qstationary.Instead of maximising or minimising Q = x T Ax subject to the constraintx T x = 1, an equivalent procedure is to extremise the functionλ(x) = xT Axx T x .◮Show that if λ(x) is stationary then x is an eigenvector of A and λ(x) is equal to thecorresponding eigenvalue.We require ∆λ(x) = 0 with respect to small variations in x. Now∆λ = 1 [(x T x) ( ∆x T Ax + x T A ∆x ) − x T Ax ( ∆x T x + x T ∆x )](x T x) 2 ( )= 2∆xT Ax x T Ax ∆x T x− 2x T x x T x x T x ,since x T A ∆x =(∆x T )Ax and x T ∆x =(∆x T )x. Thus∆λ = 2x T x ∆xT [Ax − λ(x)x].Hence, if ∆λ =0thenAx = λ(x)x, i.e.x is an eigenvector of A with eigenvalue λ(x). ◭Thus the eigenvalues of a symmetric matrix A are the values of the functionλ(x) = xT Axx T xat its stationary points. The eigenvectors of A lie along those directions in spacefor which the quadratic form Q = x T Ax has stationary values, given a fixedmagnitude for the vector x. Similar results hold for Hermitian matrices.291