Mathematical Methods for Physics and Engineering : A Comprehensive Guide

MATRICES AND VECTOR SPACES

This set of equations is also triangular, and we easily find the solution

x 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 determinant

ofAfrom itsLUdecomposition. To find the inverseA−^1 , one solves the system

of equationsAx=brepeatedly for theNdifferent RHS column matricesb=ei,

i=1, 2 ,...,N,whereeiis the column matrix with itsith element equal to unity

and the others equal to zero. The solutionxin each case gives the corresponding

column ofA−^1. Evaluation of the determinant|A|is much simpler. From (8.125),

we have

|A|=|LU|=|L||U|. (8.127)

SinceLandUare triangular, however, we see from (8.64) that their determinants

are equal to the products of their diagonal elements. SinceLii= 1 for alli,we

thus find

|A|=U 11 U 22 ···UNN=

∏N

i=1

Uii.

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 matrixAis symmetric and positive semi-definite

then we can decompose it as

A=LL†, (8.128)

whereLis a lower triangular matrix whose diagonal elements arenot, in general,

equal to unity. This is known as aCholesky decomposition(in the special case

whereAis real, the decomposition becomesA=LLT). The reason that we cannot

set the diagonal elements ofLequal to unity in this case is that we require the

same number of independent elements inLas inA. The requirement that the

matrix be positive semi-definite is easily derived by considering the Hermitian

form (or quadratic form in the real case)

x†Ax=x†LL†x=(L†x)†(L†x).

Denoting the column matrixL†xbyy, we see that the last term on the RHS

isy†y, which must be greater than or equal to zero. Thus, we requirex†Ax≥ 0

for any arbitrary column matrixx,andsoAmust be positive semi-definite (see

section 8.17).

We recall that the requirement that a matrix be positive semi-definite is equiv-

alent to demanding that all the eigenvalues ofAare positive or zero. If one of

the eigenvalues ofAis zero, however, then from (8.103) we have|A|= 0 and soA

issingular. Thus, ifAis a non-singular matrix, it must bepositive definite(rather