12.12. THE MOORE PENROSE INVERSE 319
and so (V ∗x)k = σ−1 (U∗y)k . Thus the first k entries of V ∗x are determined. In order tomake |V ∗x| as small as possible, the remaining n− k entries should equal zero. Therefore,
V ∗x =
((V ∗x)k
0
)=
(σ−1 (U∗y)k
0
)=
(σ−1 0
0 0
)U∗y
and so
x = V
(σ−1 0
0 0
)U∗y ≡ A+y ■
Lemma 12.12.3 The matrix A+ satisfies the following conditions.
AA+A = A, A+AA+ = A+, A+A and AA+ are Hermitian. (12.19)
Proof: This is routine. Recall
A = U
(σ 0
0 0
)V ∗
and
A+ = V
(σ−1 0
0 0
)U∗
so you just plug in and verify it works. ■A much more interesting observation is that A+ is characterized as being the unique
matrix which satisfies 12.19. This is the content of the following Theorem. The conditionsare sometimes called the Penrose conditions.
Theorem 12.12.4 Let A be an m × n matrix. Then a matrix A0, is the Moore Penroseinverse of A if and only if A0 satisfies
AA0A = A, A0AA0 = A0, A0A and AA0 are Hermitian. (12.20)
Proof: From the above lemma, the Moore Penrose inverse satisfies 12.20. Suppose thenthat A0 satisfies 12.20. It is necessary to verify that A0 = A+. Recall that from the singularvalue decomposition, there exist unitary matrices, U and V such that
U∗AV = Σ ≡
(σ 0
0 0
), A = UΣV ∗.
Recall that
A+ = V
(σ−1 0
0 0
)U∗
Let
A0 = V
(P Q
R S
)U∗ (12.21)
where P is r × r, the same size as the diagonal matrix composed of the singular values onthe main diagonal.
Next use the first equation of 12.20 to write
A︷ ︸︸ ︷UΣV ∗
A0︷ ︸︸ ︷V
(P Q
R S
)U∗
A︷ ︸︸ ︷UΣV ∗ =
A︷ ︸︸ ︷UΣV ∗.