Mathematical Methods for Physics and Engineering : A Comprehensive Guide

MATRICES AND VECTOR SPACES

that is,Sij=(xj)i. ThereforeA′is given by

(S−^1 AS)ij=

∑

k

∑

l

(S−^1 )ikAklSlj

=

∑

k

∑

l

(S−^1 )ikAkl(xj)l

=

∑

k

(S−^1 )ikλj(xj)k

=

∑

k

λj(S−^1 )ikSkj=λjδij.

So the matrixA′is diagonal with the eigenvalues ofAas the diagonal elements,

i.e.

A′=













λ 1 0 ··· 0

0 λ 2

..

.

..

.

..

. 0
0 ··· 0 λN













.

Therefore, given a matrixA, if we construct the matrixSthat has the eigen-

vectors ofAas its columns then the matrixA′=S−^1 ASis diagonal and has the

eigenvalues ofAas its diagonal elements. Since we requireSto be non-singular

(|S|= 0), theNeigenvectors ofAmust be linearly independent and form a basis

for theN-dimensional vector space. It may be shown thatany matrix with distinct

eigenvaluescan be diagonalised by this procedure. If, however, a general square

matrix has degenerate eigenvalues then it may, or may not, haveNlinearly

independent eigenvectors. If it does not then itcannotbe diagonalised.

For normal matrices (which include Hermitian, anti-Hermitian and unitary

matrices) theNeigenvectors are indeed linearly independent. Moreover, when

normalised, these eigenvectors form anorthonormalset (or can be made to do

so). Therefore the matrixSwith these normalised eigenvectors as columns, i.e.

whose elements areSij=(xj)i,hastheproperty

(S†S)ij=

∑

k

(S†)ik(S)kj=

∑

k

Ski∗Skj=

∑

k

(xi)∗k(xj)k=(xi)

†

xj=δij.

HenceSis unitary (S−^1 =S†) and the original matrixAcan be diagonalised by

A′=S−^1 AS=S†AS.

Therefore, any normal matrixAcan be diagonalised by a similarity transformation

using aunitarytransformation matrixS.