How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 9 results · all verified · 7 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Moore--Penrose Pseudoinverse and Regularised Least Squares: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A diagonal rank-deficient matrix makes $A^{+}$ , $A A^{+}$ , and $A^{+} A$ completely explicit

Example

Let

$A = [\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array}] .$

Then

$A^{+} = [\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 0 \end{array}], A A^{+} = A^{+} A = [\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}] .$

So both $A A^{+}$ and $A^{+} A$ are the coordinate projection onto the first axis.

Facts & Assumptions

Given: The displayed diagonal matrix $A$ .

[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]

$A A^{+}$ 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 $im A^{*}$ ).

Verification

technique · computation

1.1

Because $A$ is already diagonal with one nonzero singular value, inverting that entry and leaving the zero entry fixed gives the candidate $A^{+} = diag (1 / 2, 0)$ .

L1algebra

2.1

Direct multiplication gives $A A^{+} = A^{+} A = diag (1, 0)$ .

step 1.1algebra

3.1

The matrix $diag (1, 0)$ is the orthogonal projection onto the first coordinate axis, which is exactly $im A = im A^{*}$ . This matches [L2].

L2step 2.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31Open item page →

An overdetermined inconsistent system has a unique minimum-norm least-squares solution given by $A^{+} b$

Example

Take

$A = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{array}], b = [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}] .$

Then $b \notin im A$ , the least-squares problem is inconsistent, and

$A^{+} b = [\begin{array}{c} \frac{1}{3} \\ \frac{1}{3} \end{array}] .$

Facts & Assumptions

Given: The matrix $A$ and right-hand side $b$ in the statement.

[L1]

Reduced QR gives the pseudoinverse of a full-column-rank matrix without forming normal equations (Reduced QR gives the full-column and full-row-rank pseudoinverse formulas without forming normal equations).

[L2]

$A^{+} b$ is the unique least-squares solution of minimum norm (For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm).

Verification

technique · computation

1.1

The columns of $A$ are independent, so $A$ has full column rank. Directly, $A^{*} A = [\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}], (A^{*} A)^{- 1} = \frac{1}{3} [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}] .$ Hence $A^{+} b = (A^{*} A)^{- 1} A^{*} b = (1 / 3, 1 / 3)^{T}$ .

L1algebra

2.1

The residual is $r = b - A A^{+} b = [\begin{array}{c} \frac{2}{3} \\ \frac{2}{3} \\ - \frac{2}{3} \end{array}],$ so $r \neq 0$ and the system is inconsistent.

step 1.1algebra

3.1

By [L2], the vector from step 1.1 is nevertheless the unique least-squares solution of minimum Euclidean norm.

L2step 1.1step 2.1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31Open item page →

An underdetermined consistent system has infinitely many exact solutions but a unique minimum-norm one

Example

Let

$A = [\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}], b = [\begin{array}{c} 1 \\ 1 \end{array}] .$

Then the exact solutions are

$x (t) = [\begin{array}{c} 1 - t \\ 1 - t \\ t \end{array}] (t \in F),$

and the minimum-norm one is $A^{+} b = (1 / 3, 1 / 3, 2 / 3)^{T}$ .

Facts & Assumptions

Given: The displayed matrix $A$ and right-hand side $b$ .

[L1]

If $A$ has full row rank, then $A^{+} = A^{*} (A A^{*})^{- 1}$ (If $A$ has full row rank, then $A^{+} = A^{*} (A A^{*})^{- 1}$ ).

[L2]

$A^{+} b$ is the unique minimum-norm least-squares solution (For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm).

Verification

technique · computation

1.1

Solving $A x = b$ gives the one-parameter family $x (t) = (1 - t, 1 - t, t)^{T}$ . Thus the system is consistent and has infinitely many exact solutions.

algebra

1.2

Here $A A^{*} = [\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}], (A A^{*})^{- 1} = \frac{1}{3} [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}],$ so [L1] gives $A^{+} b = A^{*} (A A^{*})^{- 1} b = [\begin{array}{c} \frac{1}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{array}] .$

L1algebra

2.1

The vector of step 1.2 equals $x (2 / 3)$ from step 1.1, so it is an exact solution. By [L2], it is the unique exact solution of minimum Euclidean norm.

L2step 1.1step 1.2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A rank-deficient matrix has an affine family of least-squares minimisers of the form $A^{+} b + (I - A^{+} A) z$

Example

Let

$A = [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}], b = [\begin{array}{c} 1 \\ 0 \end{array}] .$

Then

$A^{+} b = [\begin{array}{c} \frac{1}{4} \\ \frac{1}{4} \end{array}], I - A^{+} A = \frac{1}{2} [\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array}],$

and every least-squares minimiser is $A^{+} b + t (1, - 1)^{T}$ .

Facts & Assumptions

Given: The rank-one matrix $A$ and the vector $b$ .

[L1]

Every least-squares minimiser has the form $A^{+} b + (I - A^{+} A) z$ (Every least-squares solution has the form $A^{+} b + (I - A^{+} A) z$ , and the same affine family specializes to exact solutions when $b \in im A$ ).

Verification

technique · computation

1.1

Since $A = 2 u u^{*}$ with $u = (1, 1)^{T} / \sqrt{2}$ , the pseudoinverse is $A^{+} = \frac{1}{2} u u^{*} = \frac{1}{4} ⁣ [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}]$ . Hence $A^{+} b = [\begin{array}{c} \frac{1}{4} \\ \frac{1}{4} \end{array}], A^{+} A = \frac{1}{2} [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}] .$

algebra

2.1

Therefore $I - A^{+} A = \frac{1}{2} [\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array}],$ whose image is the kernel line $span {(1, - 1)^{T}}$ .

step 1.1algebra

3.1

By [L1], every least-squares minimiser equals $A^{+} b + (I - A^{+} A) z$ , and step 2.1 shows that this is exactly the affine line $A^{+} b + t (1, - 1)^{T}$ .

L1step 2.1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A full-column-rank pseudoinverse can be read off directly from a reduced QR factorisation

Example

Let

$A = [\begin{array}{cc} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array}] = \hat{Q} \hat{R},$

where

$\hat{Q} = [\begin{array}{cc} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{2}{\sqrt{6}} \end{array}], \hat{R} = [\begin{array}{cc} \sqrt{2} & \frac{1}{\sqrt{2}} \\ 0 & \frac{\sqrt{6}}{2} \end{array}] .$

Then

$A^{+} = {\hat{R}}^{- 1} {\hat{Q}}^{*} = [\begin{array}{ccc} \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \end{array}] .$

Facts & Assumptions

Given: The displayed reduced QR factorisation of $A$ .

[L1]

For full column rank, reduced QR gives $A^{+} = {\hat{R}}^{- 1} {\hat{Q}}^{*}$ (Reduced QR gives the full-column and full-row-rank pseudoinverse formulas without forming normal equations).

Verification

technique · computation

1.1

Direct multiplication shows ${\hat{Q}}^{*} \hat{Q} = I$ and $\hat{Q} \hat{R} = A$ , so the displayed matrices are indeed a reduced QR factorisation.

algebra

2.1

The inverse of $\hat{R}$ is ${\hat{R}}^{- 1} = [\begin{array}{cc} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ 0 & \frac{2}{\sqrt{6}} \end{array}] .$ Therefore [L1] gives the displayed matrix for $A^{+}$ .

L1step 1.1algebra

3.1

Multiplying $A^{+} A$ gives the $2 \times 2$ identity, confirming that the matrix from step 2.1 is the full-column-rank pseudoinverse.

step 2.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Ridge regularisation damps the small-singular-value directions in a diagonal inverse problem

Example

Let

$A = diag (1, 10^{- 2}), b = [\begin{array}{c} 1 \\ 1 \end{array}], λ = 10^{- 2} .$

Then the pseudoinverse gives $(1, 100)^{T}$ , whereas the Tikhonov solution is

$x_{λ} = [\begin{array}{c} \frac{1}{1 + λ} \\ \frac{10^{- 2}}{10^{- 4} + λ} \end{array}] = [\begin{array}{c} \frac{100}{101} \\ \frac{100}{101} \end{array}] .$

Facts & Assumptions

Given: The diagonal matrix $A$ , the vector $b$ , and the parameter $λ = 10^{- 2}$ .

[L1]

Tikhonov regularisation scales the $i$ th singular direction by $σ_{i} / (σ_{i}^{2} + λ)$ (Tikhonov regularisation scales each singular component by the filter factor $σ / (σ^{2} + λ)$ ).

Verification

technique · computation

1.1

The singular values are $1$ and $10^{- 2}$ , so the unregularised pseudoinverse multiplies the second component of $b$ by $10^{2}$ , producing the large coefficient $100$ .

algebra

1.2

By [L1], the Tikhonov filter factors are $\frac{1}{1 + λ} = \frac{100}{101}, \frac{10^{- 2}}{10^{- 4} + λ} = \frac{100}{101},$ so the regularised solution is the displayed vector.

L1algebra

2.1

Step 1.2 shows that the small singular direction is strongly damped: the unstable coefficient $100$ is replaced by $100 / 101$ .

step 1.1step 1.2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Truncated SVD denoising on a diagonal problem keeps the declared large singular directions and discards the noisy ones

Example

Let

$A = diag (4, 10^{- 2}), b = [\begin{array}{c} 8 \\ 1 \end{array}], τ = 10^{- 1} .$

Then the truncated pseudoinverse keeps the first singular direction and drops the second, so

$A_{τ}^{+} b = [\begin{array}{c} 2 \\ 0 \end{array}] .$

Facts & Assumptions

Given: The diagonal matrix $A$ , the data vector $b$ , and the threshold $τ = 10^{- 1}$ .

[L1]

The truncated singular-value pseudoinverse keeps the reciprocal $1 / σ_{i}$ exactly when $σ_{i} > τ$ (The truncated singular-value pseudoinverse obtained by discarding singular values below a declared numerical-rank threshold).

[L2]

Truncated SVD is the hard spectral filter among the regularisation methods (Truncated SVD and Tikhonov regularisation act as hard and smooth spectral filters on the singular components).

Verification

technique · computation

1.1

The singular values of $A$ are $4$ and $10^{- 2}$ . Since $4 > τ$ but $10^{- 2} \leq τ$ , [L1] gives $A_{τ}^{+} = diag ⁣ (\frac{1}{4}, 0) .$

L1algebra

2.1

Multiplying by $b$ yields $A_{τ}^{+} b = diag ⁣ (\frac{1}{4}, 0) [\begin{array}{c} 8 \\ 1 \end{array}] = [\begin{array}{c} 2 \\ 0 \end{array}] .$

step 1.1algebra

3.1

The second data component is discarded entirely because it lies in the singular direction below the declared threshold, exactly as [L2] predicts.

L2step 2.1∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

For arbitrary matrices, $(A B)^{+} = B^{+} A^{+}$

Statement

False claim: for arbitrary compatible matrices $A$ and $B$ ,

(A B)^{+} = B^{+} A^{+} .

Facts & Assumptions

Given: The matrices

$A = [\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array}], B = [\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array}] .$

[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).

Refutation

technique · direct

1.1

The three matrices are rank one: $A B = [\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array}], A^{+} = \frac{1}{2} [\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array}], B^{+} = \frac{1}{2} [\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array}] .$

L1algebra

2.1

Therefore $(A B)^{+} = [\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 0 \end{array}], B^{+} A^{+} = \frac{1}{4} [\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}] .$

step 1.1algebra

3.1

The two matrices in step 2.1 are different, so the displayed matrices refute the claim $(A B)^{+} = B^{+} A^{+}$ .

step 2.1∎

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

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∎

Sources

Roger Penrose, A generalized inverse for matrices

The Moore--Penrose Pseudoinverse and Regularised Least Squares: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A diagonal rank-deficient matrix makes A+A+, AA+AA+, and A+AA+A completely explicit

Example

Facts & Assumptions

Verification

An overdetermined inconsistent system has a unique minimum-norm least-squares solution given by A+bA+b

Example

Facts & Assumptions

Verification

An underdetermined consistent system has infinitely many exact solutions but a unique minimum-norm one

Example

Facts & Assumptions

Verification

A rank-deficient matrix has an affine family of least-squares minimisers of the form A+b+(I−A+A)zA+b+(I−A+A)z

Example

Facts & Assumptions

Verification

A full-column-rank pseudoinverse can be read off directly from a reduced QR factorisation

Example

Facts & Assumptions

Verification

Ridge regularisation damps the small-singular-value directions in a diagonal inverse problem

Example

Facts & Assumptions

Verification

Truncated SVD denoising on a diagonal problem keeps the declared large singular directions and discards the noisy ones

Example

Facts & Assumptions

Verification

For arbitrary matrices, (AB)+=B+A+(AB)+=B+A+

Statement

Facts & Assumptions

Refutation

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

Statement refuted

Facts & Assumptions

Counterexample

Sources

A diagonal rank-deficient matrix makes $A^{+}$ , $A A^{+}$ , and $A^{+} A$ completely explicit

An overdetermined inconsistent system has a unique minimum-norm least-squares solution given by $A^{+} b$

A rank-deficient matrix has an affine family of least-squares minimisers of the form $A^{+} b + (I - A^{+} A) z$

For arbitrary matrices, $(A B)^{+} = B^{+} A^{+}$

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