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Full-column-rank matrix least squares has a unique minimizer

Statement

Let $p, r, s \geq 1$ , let $M \in F^{p \times r}$ have full column rank, and let $Y \in F^{p \times s}$ . Among all $W \in F^{r \times s}$ , the function $W \mapsto ∥ Y - M W ∥_{F}$ has the unique minimizer $W_{*} = (M^{*} M)^{- 1} M^{*} Y .$

Facts & Assumptions

Given: A full-column-rank matrix $M$ and a target matrix $Y$ with the same row count.

[F1]

Finite tensor arrays and the Frobenius inner product: the Frobenius inner product is the sum of conjugate-entry times entry, so orthogonal matrix residuals have squared norms that add.

Proof

1.1givenconstructalgebra

The Gram matrix $G = M^{*} M$ is invertible: if $G x = 0$ , then $0 = x^{*} G x = ∥ M x ∥_{2}^{2}$ , so $M x = 0$ ; full column rank forces $x = 0$ . Thus the square map $G : F^{r} \to F^{r}$ is injective and therefore bijective in finite dimension. Define $W_{*} = G^{- 1} M^{*} Y$ .

2.1F1step 1.1algebra

Set $R = Y - M W_{*}$ . Then $M^{*} R = M^{*} Y - M^{*} M W_{*} = 0$ . For every $W$ , write $Y - M W = R + M (W_{*} - W)$ . The two summands are Frobenius-orthogonal because $M^{*} R = 0$ , so $∥ Y - M W ∥_{F}^{2} = ∥ R ∥_{F}^{2} + ∥ M (W - W_{*}) ∥_{F}^{2} \geq ∥ R ∥_{F}^{2}$ .

3.1givenstep 2.1algebra∎

Equality in step 2.1 requires $M (W - W_{*}) = 0$ . Full column rank implies each column of $W - W_{*}$ is zero, hence $W = W_{*}$ . Therefore $W_{*}$ is the unique minimizer.

Depends on

Finite tensor arrays and the Frobenius inner product

Used by

Budgeted CP alternating least squares Algorithm
Exact CP-ALS block updates do not increase the objective Theorem

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Sources

ACM 270-1, Complete Lecture Notes, Caltech, Spring 2019, Lecture 13, §4.3 (standard reference, not scraped)
Tamara G. Kolda and Brett W. Bader, Tensor Decompositions and Applications, SIAM Review 51(3) (2009), §3.4 (standard reference, not scraped)