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

Statement

Let $A \in M_{m \times n} (F)$ and $b \in F^{m}$ . A vector $x$ is a least-squares solution of $A x = b$ if and only if

$x = A^{+} b + (I - A^{+} A) z$

for some $z \in F^{n}$ . If $b \in im A$ , then the same family is exactly the full solution set of $A x = b$ .

Facts & Assumptions

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

[L1]

$A^{+} b$ is a least-squares minimiser, and every least-squares minimiser differs from it by an element of $\ker A$ (For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm, For a linear map $T : V \to W$ between finite-dimensional inner-product spaces, $x$ minimises $∥ T x - b ∥$ if and only if $T^{*} (T x - b) = 0$ , equivalently $T^{*} T x = T^{*} b$ ; minimisers exist and any two differ by an element of $\ker T$ ).

[L2]

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

[L3]

$im A^{*} = (\ker A)^{⊥}$ ( $\ker T^{*} = (im T)^{⊥}$ and $im T^{*} = (\ker T)^{⊥}$ in finite dimension).

Proof

technique · direct

1.1

By [L2] and [L3], $A^{+} A$ is the orthogonal projection onto $(\ker A)^{⊥}$ . Therefore $I - A^{+} A$ is the orthogonal projection onto $\ker A$ . In particular, its image is $\ker A$ and it vanishes on $(\ker A)^{⊥}$ .

L2L3algebra

2.1

By [L1], every least-squares minimiser has the form $A^{+} b + u$ with $u \in \ker A$ . Since step 1.1 says every $u \in \ker A$ equals $(I - A^{+} A) z$ for some $z$ , every least-squares minimiser has the stated form.

L1step 1.1

2.2

Conversely, if $x = A^{+} b + (I - A^{+} A) z$ , then step 1.1 gives $(I - A^{+} A) z \in \ker A$ . Hence $x$ differs from the least-squares minimiser $A^{+} b$ by a kernel vector, so [L1] implies that $x$ is again a least-squares minimiser.

L1step 1.1algebra

3.1

If $b \in im A$ , then the least-squares residual can be $0$ . Thus the least-squares minimisers are exactly the exact solutions of $A x = b$ , and steps 2.1-2.1 identify that exact solution set with the same affine family.

L1step 2.1step 2.2

4.1

Steps 2.1 and 2.2 prove the if-and-only-if description, and step 3.1 gives the consistent specialisation.

step 2.1step 2.2step 3.1∎

Depends on

Used by

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

Dependency tree · two levels

15 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

Gene H. Golub, Least squares, singular values and matrix approximations (standard reference, not scraped)

Every least-squares solution has the form A+b+(I−A+A)zA+b+(I−A+A)z, and the same affine family specializes to exact solutions when b∈im⁡Ab∈imA