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

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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∎

Depends on

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

Used by

Nothing in the library uses this result yet.

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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