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 matrixA=
22 2 2
17
101
10 −17
10 −1
10
3
59
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 haveS=
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