Mathematical Methods for Physics and Engineering - Matematica.NET

MATRICES AND VECTOR SPACESThis set of equations is also triangular, and we easily find the solutionx 1 =2, x 2 = −3, x 3 =4,which agrees with the result found above by direct inversion. ◭We note, in passing, that one can calculate both the inverse and the determinantof A from its LU decomposition. To find the inverse A −1 , one solves the systemof equations Ax = b repeatedly for the N different RHS column matrices b = e i ,i =1, 2,...,N,wheree i is the column matrix with its ith element equal to unityand the others equal to zero. The solution x in each case gives the correspondingcolumn of A −1 . Evaluation of the determinant |A| is much simpler. From (8.125),we have|A| = |LU| = |L||U|. (8.127)Since L and U are triangular, however, we see from (8.64) that their determinantsare equal to the products of their diagonal elements. Since L ii = 1 for all i, wethus findN∏|A| = U 11 U 22 ···U NN = U ii .As an illustration, in the above example we find |A| = (2)(−4)(−11/8) = 11,which, as it must, agrees with our earlier calculation (8.58).Finally, we note that if the matrix A is symmetric and positive semi-definitethen we can decompose it asi=1A = LL † , (8.128)where L is a lower triangular matrix whose diagonal elements are not, in general,equal to unity. This is known as a Cholesky decomposition (in the special casewhere A is real, the decomposition becomes A = LL T ). The reason that we cannotset the diagonal elements of L equal to unity in this case is that we require thesame number of independent elements in L as in A. The requirement that thematrix be positive semi-definite is easily derived by considering the Hermitianform (or quadratic form in the real case)x † Ax = x † LL † x =(L † x) † (L † x).Denoting the column matrix L † x by y, we see that the last term on the RHSis y † y, which must be greater than or equal to zero. Thus, we require x † Ax ≥ 0for any arbitrary column matrix x, andsoA must be positive semi-definite (seesection 8.17).We recall that the requirement that a matrix be positive semi-definite is equivalentto demanding that all the eigenvalues of A are positive or zero. If one ofthe eigenvalues of A is zero, however, then from (8.103) we have |A| = 0 and so Ais singular. Thus, if A is a non-singular matrix, it must be positive definite (rather298