PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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Pseudoinversion is involutive, commutes with adjoints, and is equivariant under unitary left and right factors

Statement

Let $A$ be a finite real or complex matrix.

$(A^{+})^{+} = A$ .
$(A^{*})^{+} = (A^{+})^{*}$ .
If $U$ and $V$ are unitary of compatible sizes, then $(U A V)^{+} = V^{*} A^{+} U^{*}$ .

Facts & Assumptions

Given: A matrix $A$ over $R$ or $C$ , and compatible unitary matrices $U$ and $V$ .

[L1]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse (Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse).

[L2]

Unitary operators preserve the inner product and satisfy $U^{*} U = I = U U^{*}$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

Proof

technique · direct

1.1L1algebra

Because $A^{+}$ satisfies the Penrose equations for $A$ , the same equations read in reverse order show that $A$ satisfies the Penrose equations for $A^{+}$ : $A^{+} A A^{+} = A^{+}, A A^{+} A = A,$ and the products $A^{+} A$ and $A A^{+}$ are already self-adjoint.

1.2L2algebra

Put $B : = V^{*} A^{+} U^{*}$ . Using [L2] and the Penrose equations for $A$ , $(U A V) B (U A V) = U A A^{+} A V = U A V,$ and similarly $B (U A V) B = B$ . The products $(U A V) B$ and $B (U A V)$ are $U (A A^{+}) U^{*}$ and $V^{*} (A^{+} A) V$ , hence self-adjoint.

2.1L1step 1.1

By uniqueness in [L1], the Moore--Penrose pseudoinverse of $A^{+}$ is $A$ . Hence $(A^{+})^{+} = A$ .

2.2step 1.1algebra

Taking adjoints of the Penrose equations for $A$ shows that $(A^{+})^{*}$ obeys $A^{*} (A^{+})^{*} A^{*} = A^{*}, (A^{+})^{*} A^{*} (A^{+})^{*} = (A^{+})^{*},$ and that $A^{*} (A^{+})^{*}$ and $(A^{+})^{*} A^{*}$ are self-adjoint.

3.1L1step 2.2

Therefore $(A^{+})^{*}$ is the Moore--Penrose pseudoinverse of $A^{*}$ , so uniqueness in [L1] gives $(A^{*})^{+} = (A^{+})^{*}$ .

4.1L1step 2.1step 3.1step 1.2∎

So $B$ is a Moore--Penrose pseudoinverse of $U A V$ , and [L1] yields $(U A V)^{+} = V^{*} A^{+} U^{*}$ . Together with steps 2.1 and 3.1, this proves the three claims.

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Roger Penrose, A generalized inverse for matrices (standard reference, not scraped)