13.18. THE MOORE PENROSE INVERSE 379

Proof: This is routine. Recall

A =U

(σ 00 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 13.35. This is the content of the following Theorem. The conditionsare sometimes called the Penrose conditions.

Theorem 13.18.4 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. (13.36)

Proof: From the above lemma, the Moore Penrose inverse satisfies 13.36. Supposethen that A0 satisfies 13.36. It is necessary to verify that A0 = A+. Recall that from thesingular value decomposition, there exist unitary matrices, U and V such that

U∗AV = Σ≡

(σ 00 0

), A =UΣV ∗.

Recall that

A+ =V

(σ−1 0

0 0

)U∗

Let

A0 =V

(P QR S

)U∗ (13.37)

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 13.36 to write

A︷ ︸︸ ︷UΣV ∗

A0︷ ︸︸ ︷V

(P QR S

)U∗

A︷ ︸︸ ︷UΣV ∗ =

A︷ ︸︸ ︷UΣV ∗.

Then multiplying both sides on the left by U∗ and on the right by V,(σ 00 0

)(P QR S

)(σ 00 0

)=

(σPσ 0

0 0

)=

(σ 00 0

)(13.38)

Therefore, P = σ−1. From the requirement that AA0 is Hermitian,

A︷ ︸︸ ︷UΣV ∗

A0︷ ︸︸ ︷V

(P QR S

)U∗ =U

(σ 00 0

)(P QR S

)U∗