LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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The minimum-norm indicator-response least-squares fit exists uniquely

Statement

Let $n, q \geq 1$ and $K \geq 2$ be finite, let $X \in R^{n \times q}$ , and let $Y \in R^{n \times K}$ be an indicator response matrix. Among all minimizers of $∥ Y - X B ∥_{F}^{2}$ over $B \in R^{q \times K}$ there is a unique one of smallest Frobenius norm. If $X = U_{r} D_{r} V_{r}^{T}$ is a thin singular-value decomposition with rank $r > 0$ , orthonormal columns in $U_{r}, V_{r}$ , and positive diagonal entries in $D_{r}$ , then this minimizer is $B_{*} = V_{r} D_{r}^{- 1} U_{r}^{T} Y$ . If $r = 0$ , then $X = 0$ and $B_{*} = 0$ . In either case the full minimizer set is ${B_{*} + N : X N = 0}$ .

Statement references

Indicator-response least-squares classification

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 $\ker T$

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition

Hastie, Tibshirani, and Friedman, ESL, §4.2, equations (4.3)–(4.6)

Stanford CS229 Spring 2022 lecture notes, §§1.2 and 2.1

Facts & Assumptions

Given: Finite real matrices $X \in R^{n \times q}$ and $Y \in R^{n \times K}$ , with $n, q \geq 1$ and $K \geq 2$ .

[F1]

The indicator-response fit minimizes $(2 n)^{- 1} ∥ Y - X B ∥_{F}^{2}$ over coefficient matrices; multiplying the objective by $2 n$ does not change its minimizers. Indicator-response least-squares classification

[F2]

For a finite-dimensional linear map, least-squares minimizers exist, satisfy the normal equation, and any two differ by an element of the kernel. 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 $\ker T$

[F3]

Every finite-dimensional real inner-product-space map has a singular-value decomposition with orthonormal input and output bases and positive singular values on its image. Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition

Proof

1.1F1F3givenalgebra

If $r = 0$ , [F3] implies $X = 0$ . By the objective in [F1], the least-squares objective is then constant in $B$ , so every matrix is a minimizer and $B_{*} = 0$ is the unique one of smallest Frobenius norm. The minimizer set is ${N : X N = 0}$ , as claimed.

1.2F3givenconstructalgebra

Suppose $r > 0$ and write the thin SVD $X = U_{r} D_{r} V_{r}^{T}$ from [F3]. Define $B_{*} = V_{r} D_{r}^{- 1} U_{r}^{T} Y$ . Orthonormality gives $X B_{*} = U_{r} U_{r}^{T} Y$ , whose $k$ th column is the orthogonal projection of the $k$ th column of $Y$ onto $im X = span (U_{r})$ .

2.1F2step 1.2

Each residual column of $Y - X B_{*}$ is therefore orthogonal to $im X$ . Applying [F2] to each column shows that every column of $B_{*}$ is a least-squares minimizer, hence $B_{*}$ minimizes the matrix objective.

2.2F2F3step 1.2

By [F2], any other matrix minimizer has the form $B_{*} + N$ with $X N = 0$ . The SVD in [F3] gives $\ker X = span (v_{r + 1}, \dots, v_{q})$ , orthogonal to the row space $span (v_{1}, \dots, v_{r})$ containing every column of $B_{*}$ .

3.1step 1.1step 2.2algebra∎

Columnwise Pythagoras now gives $∥ B_{*} + N ∥_{F}^{2} = ∥ B_{*} ∥_{F}^{2} + ∥ N ∥_{F}^{2}$ . Thus $B_{*}$ has the smallest Frobenius norm among all minimizers, and equality is possible only for $N = 0$ . Together with step 1.1, this proves existence and uniqueness also in rank zero.

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by Indicator-response least-squares classification.

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Sources

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing (standard reference, not scraped)