CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The Moore--Penrose pseudoinverse exchanges image and adjoint-image, and exchanges kernel and adjoint-kernel

Statement

For every finite real or complex matrix $A$ ,

$im A^{+} = im A^{*}, im (A^{+})^{*} = im A,$

and

$\ker A^{+} = \ker A^{*}, \ker (A^{+})^{*} = \ker A .$

Facts & Assumptions

Given: A finite real or complex matrix $A$ .

[L1]

$A A^{+}$ and $A^{+} A$ are the orthogonal projections onto $im A$ and $im A^{*}$ ( $A A^{+}$ and $A^{+} A$ are the orthogonal projections onto $im A$ and $im A^{*}$ ).

[L2]

Pseudoinversion is involutive and adjoint-compatible: $(A^{+})^{+} = A$ and $(A^{+})^{*} = (A^{*})^{+}$ (Pseudoinversion is involutive, commutes with adjoints, and is equivariant under unitary left and right factors).

[L3]

For every finite-dimensional operator, $\ker T^{*} = (im T)^{⊥}$ and $\ker T = (im T^{*})^{⊥}$ ( $\ker T^{*} = (im T)^{⊥}$ and $im T^{*} = (\ker T)^{⊥}$ in finite dimension).

Proof

technique · direct

1.1L1L2

Apply [L1] to $A$ : $A^{+} A$ is the orthogonal projection onto $im A^{*}$ . Apply [L1] again to $A^{+}$ and use [L2] to identify $(A^{+})^{+} = A$ ; then $A^{+} A$ is also the orthogonal projection onto $im A^{+}$ . Orthogonal projections onto a given space are unique, so $im A^{+} = im A^{*}$ .

2.1L1L2step 1.1

Using [L2] in the same way, $A A^{+}$ is both the orthogonal projection onto $im A$ and the orthogonal projection onto $im (A^{+})^{*}$ . Hence $im (A^{+})^{*} = im A$ .

2.2L3step 1.1algebra

By [L3] and step 1.1, $\ker A = (im A^{*})^{⊥} = (im A^{+})^{⊥} = \ker (A^{+})^{*} .$

3.1L3step 2.1algebra

By [L3] and step 2.1, $\ker A^{*} = (im A)^{⊥} = (im (A^{+})^{*})^{⊥} = \ker A^{+} .$

4.1step 1.1step 2.1step 3.1step 2.2∎

Steps 1.1, 2.1, 3.1, and 2.2 give the image and kernel identities.

Depends on

Used by

Dependency tree · two levels

12 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)
Stanford CS205L, Unit 11: Zero Singular Values (standard reference, not scraped)