Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

8.18 SIMULTANEOUS LINEAR EQUATIONS


Show that, fori=1, 2 ,...,p,Avi=siuiandA†ui=sivi,wherep=min(M, N).

Post-multiplying both sides of (8.131) byV, and using the fact thatVis unitary, we obtain


AV=US.

Since the columns ofVandUconsist of the vectorsviandujrespectively andShas only
diagonal non-zero elements, we find immediately that, fori=1, 2 ,...,p,


Avi=siui. (8.134)

Moreover, we note thatAvi=0fori=p+1,p+2,...,N.
Taking the Hermitian conjugate of both sides of (8.131) and post-multiplying byU,we
obtain


A†U=VS†=VST,

where we have used the fact thatUis unitary andSis real. We then see immediately that,
fori=1, 2 ,...,p,


A†ui=sivi. (8.135)

We also note thatA†ui=0fori=p+1,p+2,...,M. Results (8.134) and (8.135) are useful
for investigating the properties of the SVD.


The decomposition (8.131) has some advantageous features for the analysis of

sets of simultaneous linear equations. These are best illustrated by writing the


decomposition (8.131) in terms of the vectorsuiandvias


A=

∑p

i=1

siui(vi)†,

wherep= min(M, N). It may be, however, that some of the singular valuessi


arezero, as a result of degeneracies in the set ofMlinear equationsAx=b.


Let us suppose that there arernon-zero singular values. Since our convention is


to arrange the singular values in order of decreasing size, the non-zero singular


values aresi,i=1, 2 ,...,r, and the zero singular values aresr+1,sr+2,...,sp.


Therefore we can writeAas


A=

∑r

i=1

siui(vi)†. (8.136)

Let us consider the action of (8.136) on an arbitrary vectorx. This is given by

Ax=

∑r

i=1

siui(vi)†x.

Since (vi)†xis just a number, we see immediately that the vectorsui,i=1, 2 ,...,r,


must span therangeof the matrixA; moreover, these vectors form an orthonor-


mal basis for the range. Further, since this subspace isr-dimensional, we have


rankA=r, i.e. the rank ofAis equal to the number of non-zero singular values.


The SVD is also useful in characterising the null space ofA. From (8.119),

we already know that the null space must have dimensionN−r;so,ifAhasr

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