CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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The family $diag (1, t)$ shows that pseudoinversion is not continuous across rank loss

Statement refuted

Refuted claim: the Moore--Penrose pseudoinverse depends continuously on a matrix everywhere in the full matrix space.

For $t \neq 0$ , let

A_{t} = diag (1, t) .

Then $A_{t} \to diag (1, 0)$ as $t \to 0$ , but $A_{t}^{+} = diag (1, t^{- 1})$ does not converge.

Facts & Assumptions

Given: The family $A_{t} = diag (1, t)$ for $t \neq 0$ .

[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).

Counterexample

technique · direct

1.1

As $t \to 0$ , the matrices $A_{t}$ converge entrywise to $A_{0} = diag (1, 0)$ .

algebra

1.2

For every $t \neq 0$ , the inverse $A_{t}^{- 1} = diag (1, t^{- 1})$ directly satisfies the four Penrose equations, so uniqueness in [L1] makes it $A_{t}^{+}$ . Hence $∥ A_{t}^{+} ∥_{2} = \max (1, ∣ t ∣^{- 1})$ .

L1algebra

2.1

Because $∣ t ∣^{- 1} \to \infty$ as $t \to 0$ , the family $(A_{t}^{+})$ is unbounded and therefore cannot converge to the finite matrix $A_{0}^{+}$ .

step 1.2algebra

3.1

Thus a convergent matrix family can have a nonconvergent pseudoinverse family, refuting global continuity.

step 1.1step 2.1∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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.

The family diag⁡(1,t)diag(1,t) shows that pseudoinversion is not continuous across rank loss