13.5. THE SINGULAR VALUE DECOMPOSITION 319
= rank
((σ 00 0
)∗)= number of singular values. ■
How could you go about computing the singular value decomposition? The proof ofexistence indicates how to do it. Here is an informal method. You have from the singularvalue decompositon,
A =U
(σ 00 0
)V ∗, A∗ =V
(σ 00 0
)U∗
Then it follows that
A∗A =V
(σ 00 0
)U∗U
(σ 00 0
)V ∗ =V
(σ2 00 0
)V ∗
and so A∗AV =V
(σ2 00 0
). Similarly, AA∗U =U
(σ2 00 0
). Therefore, you would
find an orthonormal basis of eigenvectors for AA∗ make them the columns of a matrix suchthat the corresponding eigenvalues are decreasing. This gives U. You could then do thesame for A∗A to get V .
Example 13.5.5 Find a singular value decomposition for the matrix
A≡
(25
√2√
5 45
√2√
5 025
√2√
5 45
√2√
5 0
)
First consider A∗A 165
325 0
325
645 0
0 0 0
What are some eigenvalues and eigenvectors? Some computing shows these are
001
,
−25
√5
15
√5
0
↔ 0,
15
√5
25
√5
0
↔ 16
Thus the matrix V is given by
V =
15
√5 − 2
5
√5 0
25
√5 1
5
√5 0
0 0 1
Next consider AA∗ =
(8 88 8
). Eigenvectors and eigenvalues are{(
− 12
√2
12
√2
)}↔ 0,
{(12
√2
12
√2
)}↔ 16