Mathematical Methods for Physics and Engineering - Matematica.NET

MATRICES AND VECTOR SPACESalso. Another, rather more general, expression that is also real is the HermitianformH(x) ≡ x † Ax, (8.112)where A is Hermitian (i.e. A † = A) and the components of x may now be complex.It is straightforward to show that H is real, sinceH ∗ =(H T ) ∗ = x † A † x = x † Ax = H.With suitable generalisation, the properties of quadratic forms apply also to Hermitianforms, but to keep the presentation simple we will restrict our discussionto quadratic forms.A special case of a quadratic (Hermitian) form is one for which Q = x T Axis greater than zero for all column matrices x. By choosing as the basis theeigenvectors of A we have Q in the formQ = λ 1 x 2 1 + λ 2 x 2 2 + λ 3 x 2 3.The requirement that Q>0 for all x means that all the eigenvalues λ i of A mustbe positive. A symmetric (Hermitian) matrix A with this property is called positivedefinite. If,instead,Q ≥ 0 for all x then it is possible that some of the eigenvaluesare zero, and A is called positive semi-definite.8.17.1 The stationary properties of the eigenvectorsConsider a quadratic form, such as Q(x) =〈x|A x〉, equation (8.105), in a fixedbasis. As the vector x is varied, through changes in its three components x 1 , x 2and x 3 , the value of the quantity Q also varies. Because of the homogeneousform of Q we may restrict any investigation of these variations to vectors of unitlength (since multiplying any vector x by any scalar k simply multiplies the valueof Q by a factor k 2 ).Of particular interest are any vectors x that make the value of the quadraticform a maximum or minimum. A necessary, but not sufficient, condition for thisis that Q is stationary with respect to small variations ∆x in x, whilst 〈x|x〉 ismaintained at a constant value (unity).In the chosen basis the quadratic form is given by Q = x T Ax and, usingLagrange undetermined multipliers to incorporate the variational constraints, weare led to seek solutions of∆[x T Ax − λ(x T x − 1)] = 0. (8.113)This may be used directly, together with the fact that (∆x T )Ax = x T A ∆x, sinceAis symmetric, to obtainAx = λx (8.114)290