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

Statement

For every matrix $A \in M_{m \times n} (F)$ over $F = R$ or $C$ , there exists a unique matrix $A^{+} \in M_{n \times m} (F)$ satisfying the four Penrose equations of The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations.

Facts & Assumptions

Given: A matrix $A \in M_{m \times n} (F)$ with $F \in {R, C}$ .

[L1]

$A$ admits a singular value decomposition $A = U Σ V^{*}$ with $U$ and $V$ unitary and $Σ$ diagonal with the nonzero singular values on the diagonal (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The Moore--Penrose pseudoinverse is defined by the four Penrose equations (The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations).

Proof

technique · direct

1.1

By [L1], after choosing singular values $σ_{1}, \dots, σ_{r} > 0$ one may write $A = U [\begin{array}{cc} diag (σ_{1}, \dots, σ_{r}) & 0 \\ 0 & 0 \end{array}] V^{*} .$

L1given

2.1

Define $Σ^{+} : = [\begin{array}{cc} diag (σ_{1}^{- 1}, \dots, σ_{r}^{- 1}) & 0 \\ 0 & 0 \end{array}], B : = V Σ^{+} U^{*} .$ Then $Σ Σ^{+} Σ = Σ$ , $Σ^{+} Σ Σ^{+} = Σ^{+}$ , and the two products $Σ Σ^{+}$ and $Σ^{+} Σ$ are diagonal with only $0$ and $1$ on the diagonal, hence are self-adjoint.

step 1.1constructalgebra

3.1

Multiplying the relations of step 2.1 by the unitary factors $U$ and $V$ shows $A B A = A, B A B = B, (A B)^{*} = A B, (B A)^{*} = B A .$ Thus $B$ is a Moore--Penrose pseudoinverse of $A$ in the sense of [L2].

L2step 2.1algebra

4.1

Let $C$ be any other Moore--Penrose pseudoinverse of $A$ , and put $D : = V^{*} C U$ . Then the same unitary transport used in step 3.1 gives $Σ D Σ = Σ, D Σ D = D,$ with $Σ D$ and $D Σ$ self-adjoint.

L2step 1.1step 3.1algebra

5.1

Write $Σ = [\begin{array}{cc} S & 0 \\ 0 & 0 \end{array}], D = [\begin{array}{cc} X & Y \\ Z & W \end{array}],$ where $S = diag (σ_{1}, \dots, σ_{r})$ . From $Σ D Σ = Σ$ one gets $S X S = S$ , hence $X = S^{- 1}$ . Because $Σ D$ is self-adjoint, its upper-right block satisfies $S Y = 0$ , so $Y = 0$ . Because $D Σ$ is self-adjoint, its lower-left block satisfies $Z S = 0$ , so $Z = 0$ . With these identities, the equation $D Σ D = D$ reduces to $[\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}] = [\begin{array}{cc} X & 0 \\ 0 & W \end{array}],$ so $W = 0$ . Hence $D = Σ^{+}$ and therefore $C = V Σ^{+} U^{*} = B$ .

step 4.1algebra

6.1

Step 3.1 gives existence and step 5.1 gives uniqueness, so every finite real or complex matrix has a unique Moore--Penrose pseudoinverse.

step 3.1step 5.1∎

Depends on

Used by

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8 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)
Andrew Stuart and Jochen Voss, Matrix Analysis and Algorithms (standard reference, not scraped)