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

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The Moore--Penrose pseudoinverse is continuous on each fixed-rank stratum and is not continuous across rank loss

Statement

Let $F \in {R, C}$ .

For fixed $m$ and $n$ , on the set of $m \times n$ matrices over $F$ of a fixed rank $r$ , the map $A \mapsto A^{+}$ is continuous.
On a full matrix space the pseudoinverse need not be continuous at a rank-deficient matrix. Already for the $2 \times 2$ path $A_{t} = diag (1, t)$ with $t \neq 0$ , $∥ A_{t}^{+} ∥_{2} \to \infty$ as $t \to 0$ .

Facts & Assumptions

Given: Real or complex matrices, with the fixed-rank and rank-loss cases as in the statement.

[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]

Every matrix admits a singular value decomposition (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L3]

Rank equals the number of nonzero singular values (The rank of a linear map is the number of its nonzero singular values).

[L4]

Closed bounded subsets of a finite-dimensional Euclidean space are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1

Let $A_{k} \to A$ with every $A_{k}$ and $A$ of rank $r$ . By [L2], choose SVDs $A_{k} = U_{k} Σ_{k} V_{k}^{*}$ and $A = U Σ V^{*}$ . Each unitary group is a closed bounded subset of its finite-dimensional matrix space, hence compact by [L4]. Therefore every subsequence of $(U_{k}, V_{k})$ has a convergent subsequence; along such a subsequence the limit still gives a singular value decomposition of $A$ .

L2L4given

1.2

For $A_{t} = diag (1, t)$ with $t \neq 0$ , [L1] gives $A_{t}^{+} = diag (1, t^{- 1})$ . Thus $∥ A_{t}^{+} ∥_{2} = \max (1, ∣ t ∣^{- 1}) \to \infty$ as $t \to 0$ .

L1algebra

2.1

By [L3], exactly the first $r$ diagonal entries of every $Σ_{k}$ and of $Σ$ are positive. Define the transposed-shape matrices $Σ_{k}^{+}$ and $Σ^{+}$ by reciprocating precisely those entries and setting all remaining entries to zero. Direct diagonal multiplication verifies the four Penrose equations, so uniqueness in [L1] gives $A_{k}^{+} = V_{k} Σ_{k}^{+} U_{k}^{*}$ and $A^{+} = V Σ^{+} U^{*}$ . Along the convergent subsequence from step 1.1 the positive singular values converge to those of $A$ , hence their reciprocals converge. Therefore $A_{k}^{+} = V_{k} Σ_{k}^{+} U_{k}^{*} \to V Σ^{+} U^{*} = A^{+} .$

L1L3step 1.1algebra

2.2

The matrices $A_{t}$ converge to $diag (1, 0)$ as $t \to 0$ , but their pseudoinverses do not stay bounded, hence cannot converge to the finite matrix $diag (1, 0)^{+}$ . Therefore pseudoinversion is not continuous across rank loss.

step 1.2algebra

3.1

Let $(A_{k_{j}}^{+})$ be any subsequence. Applying the compactness argument of step 1.1 to its SVD factors produces a further subsequence to which step 2.1 applies, so that further subsequence converges to $A^{+}$ . If $(A_{k}^{+})$ did not converge to $A^{+}$ , some $ε > 0$ and a subsequence would satisfy $∥ A_{k_{j}}^{+} - A^{+} ∥_{2} \geq ε$ for every $j$ , contradicting the further subsequence just obtained. Thus $A_{k}^{+} \to A^{+}$ , proving continuity on the rank- $r$ stratum.

step 1.1step 2.1contradiction

4.1

Steps 3.1 and 2.2 prove the two claims.

step 3.1step 2.2∎

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Sources

Roger Penrose, A generalized inverse for matrices (standard reference, not scraped)
Stanford CS205L, Unit 11: Zero Singular Values (standard reference, not scraped)