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

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

Statement

Let $F \in {R, C}$ , let $A \in M_{m \times n} (F)$ , and let $b \in F^{m}$ . Then $A^{+} b$ is a least-squares solution of $A x = b$ , and among all least-squares solutions it is the unique one of minimum Euclidean norm.

Facts & Assumptions

Given: A scalar field $F \in {R, C}$ , a matrix $A \in M_{m \times n} (F)$ , and a right-hand side $b \in F^{m}$ .

[L1]

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

[L2]

$im A^{+} = im A^{*}$ (The Moore--Penrose pseudoinverse exchanges image and adjoint-image, and exchanges kernel and adjoint-kernel).

[L3]

Least-squares minimisers are exactly the solutions of $A^{*} (A x - b) = 0$ , and any two minimisers differ by an element of $\ker A$ (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 $\ker T$ ).

[L4]

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

[L5]

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

Proof

technique · direct

1.1L1L3L4

By [L1], $A A^{+} b$ is the orthogonal projection of $b$ onto $im A$ . Therefore the residual $b - A A^{+} b$ lies in $(im A)^{⊥} = \ker A^{*}$ by [L4], so $A^{*} (A A^{+} b - b) = 0$ and [L3] shows that $A^{+} b$ is a least-squares minimiser.

1.2L2L5algebra

By [L2] and [L5], $A^{+} b \in im A^{+} = im A^{*} = (\ker A)^{⊥}$ . Thus $A^{+} b$ is orthogonal to every vector in $\ker A$ .

2.1L3step 1.2algebra

Let $x$ be any least-squares minimiser. By [L3], $x - A^{+} b \in \ker A$ , so $x = A^{+} b + z$ for some $z \in \ker A$ . Step 1.2 then gives $∥ x ∥_{2}^{2} = ∥ A^{+} b ∥_{2}^{2} + ∥ z ∥_{2}^{2} \geq ∥ A^{+} b ∥_{2}^{2} .$

3.1step 2.1algebra

Equality in step 2.1 holds only when $z = 0$ , so the least-squares minimiser of minimum Euclidean norm is unique and equals $A^{+} b$ .

4.1step 1.1step 2.1step 3.1∎

Step 1.1 proves the least-squares claim and steps 2.1 and 3.1 prove the minimum-norm claim.

Depends on

Used by

Dependency tree · two levels

14 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)
Stanford CS205L, Unit 11: Zero Singular Values (standard reference, not scraped)

For every right-hand side bb, A+bA+b is the unique least-squares solution of minimum Euclidean norm