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The Moore-Penrose products are orthogonal projectors

Statement

Let $M \in F^{p \times q}$ be any finite real or complex matrix, and let $M^{†}$ be its unique Moore-Penrose inverse. Then $P = M M^{†}$ and $Q = M^{†} M$ are self-adjoint idempotents. The range of $P$ is $Ran (M)$ , and the range of $Q$ is $Ran (M^{†})$ . Thus both are orthogonal projections. If $M$ has full row rank, then $M M^{†} = I_{p}$ .

Facts & Assumptions

Given: A finite matrix $M \in F^{p \times q}$ over $F \in {R, C}$ .

[F1]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: $M^{†}$ satisfies $M M^{†} M = M$ , $M^{†} M M^{†} = M^{†}$ , and both $M M^{†}$ and $M^{†} M$ are self-adjoint.

[F2]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite real or complex matrix has a unique matrix satisfying the four Moore-Penrose equations.

Proof

1.1F1F2algebra

By [F1], $P^{*} = P$ and $Q^{*} = Q$ . Also $P^{2} = M M^{†} M M^{†} = M (M^{†} M M^{†}) = M M^{†} = P$ , and $Q^{2} = M^{†} M M^{†} M = (M^{†} M M^{†}) M = M^{†} M = Q$ . Hence both products are self-adjoint idempotents.

2.1F1step 1.1

The range of $P$ is contained in $Ran (M)$ because $P = M M^{†}$ , while $P M = M$ by [F1], so $Ran (M) \subseteq Ran (P)$ . Thus $Ran (P) = Ran (M)$ . The range of $Q$ is contained in $Ran (M^{†})$ , and $Q M^{†} = M^{†}$ by [F1], so these ranges are equal.

3.1step 1.1step 2.1algebra∎

A self-adjoint idempotent is the orthogonal projection onto its range: each vector $x$ decomposes as $P x + (I - P) x$ , the first term lies in the range, the second lies in the kernel, and self-adjointness makes the range orthogonal to the kernel. The same argument applies to $Q$ . If $M$ has full row rank, $Ran (M) = F^{p}$ , so step 2.1 gives $Ran (P) = F^{p}$ and therefore $P = I_{p}$ .

Depends on

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Sources

Per-Gunnar Martinsson and Joel A. Tropp, Randomized Numerical Linear Algebra: Foundations & Algorithms, §§13.1–13.2 (standard reference, not scraped)
Alex Gittens and Michael W. Mahoney, Revisiting the Nyström Method, §4.1 (standard reference, not scraped)