Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

MATRICES AND VECTOR SPACES


non-zero singular valuessi,i=1, 2 ,...,r, then from the worked example above


we have


Avi=0 fori=r+1,r+2,...,N.

Thus, theN−rvectorsvi,i=r+1,r+2,...,N, form an orthonormal basis for


the null space ofA.


Find the singular value decompostion of the matrix

A=





22 2 2


17
10

1
10 −

17
10 −

1
10
3
5

9
5 −

3
5 −

9
5



. (8.137)


The matrixAhas dimension 3×4(i.e.M=3,N=4),andsowemayconstructfrom
it the 3×3matrixAA†and the 4×4matrixA†A(in fact, sinceAis real, the Hermitian
conjugates are just transposes). We begin by finding the eigenvaluesλiand eigenvectorsui
of the smaller matrixAA†. This matrix is easily found to be given by


AA†=




16 0 0


(^0295125)
(^0125365)



,


and its characteristic equation reads

∣∣

∣∣


16 −λ 00
0 295 −λ^125
0 125 365 −λ


∣∣



∣∣=(16−λ)(36−^13 λ+λ

(^2) )=0.
Thus, the eigenvalues areλ 1 = 16,λ 2 =9,λ 3 = 4. Since the singular values ofAare given
bysi=



λiand the matrixSin (8.131) has the same dimensions asA, we have

S=




4000


0300


0020



, (8.138)


where we have arranged the singular values inorder of decreasing size. Now the matrixU
has as its columns the normalised eigenvectorsuiof the 3×3matrixAA†. These normalised
eigenvectors correspond to the eigenvalues ofAA†as follows:


λ 1 =16 ⇒ u^1 =(100)T
λ 2 =9 ⇒ u^2 =(0^3545 )T
λ 3 =4 ⇒ u^3 =(0 −^4535 )T,

and so we obtain the matrix


U=





10 0


0 35 −^45


(^04535)




. (8.139)


The columns of the matrixVin (8.131) are the normalised eigenvectors of the 4× 4
matrixA†A, which is given by


A†A=


1


4





29 21 3 11


21 29 11 3


3112921


11 3 21 29




.

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