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The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition

Statement

Let $T : V \to W$ be a linear map between finite-dimensional real or complex inner product spaces, let

$T v = \sum_{j = 1}^{r} s_{j} ⟨ v, e_{j} ⟩ f_{j}$

be a singular value decomposition with $s_{1} \geq \dots \geq s_{r} > 0$ , and fix an integer $k$ with $0 \leq k < \dim V$ . Define

$T_{k} v : = \sum_{j = 1}^{\min (k, r)} s_{j} ⟨ v, e_{j} ⟩ f_{j} .$

Then $rank T_{k} \leq k$ , one has

$∥ T - T_{k} ∥ = s_{k + 1}$

with the convention $s_{k + 1} = 0$ when $k \geq r$ , and every linear map $A : V \to W$ with $rank A \leq k$ satisfies

$∥ T - A ∥ \geq s_{k + 1} .$

Facts & Assumptions

Given: A linear map $T : V \to W$ between finite-dimensional real or complex inner product spaces and the rank- $k$ truncation $T_{k}$ above.

[L1]

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

[L2]

The operator norm equals the largest singular value (The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector).

[L3]

Rank-nullity holds for linear maps with finite-dimensional domain (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L4]

A linear subspace of a finite-dimensional vector space cannot have larger dimension than the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1L1L2algebra

The image of $T_{k}$ is contained in $span (f_{1}, \dots, f_{k})$ , so $rank T_{k} \leq k$ . When $k = 0$ , the defining sum for $T_{k}$ is empty and $T_{k} = 0$ ; in general $(T - T_{k}) v = \sum_{j > k} s_{j} ⟨ v, e_{j} ⟩ f_{j}$ , which is again an SVD, so [L2] gives $∥ T - T_{k} ∥ = s_{k + 1}$ , with value $0$ when $k \geq r$ .

1.2L3L4algebra

Let $A : V \to W$ have rank at most $k$ , and put $E : = span (e_{1}, \dots, e_{k + 1})$ . If $A ∣_{E}$ were injective, then [L3] applied to $A ∣_{E} : E \to W$ would force $\dim im (A ∣_{E}) = k + 1$ , but $im (A ∣_{E}) \subseteq im A$ and [L4] gives $\dim im A \leq k$ , contradiction. Hence some unit vector $v \in E$ satisfies $A v = 0$ .

2.1step 1.2algebra∎

Write that unit vector as $v = \sum_{j = 1}^{k + 1} c_{j} e_{j}$ . Then $∥ T v ∥^{2} = \sum_{j = 1}^{k + 1} s_{j}^{2} ∣ c_{j} ∣^{2} \geq s_{k + 1}^{2} \sum_{j = 1}^{k + 1} ∣ c_{j} ∣^{2} = s_{k + 1}^{2}$ . Because $A v = 0$ , one has $∥ T - A ∥ \geq ∥ (T - A) v ∥ = ∥ T v ∥ \geq s_{k + 1}$ . Combined with step 1.1, this proves that $T_{k}$ is a best rank-at-most- $k$ approximation in operator norm.

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Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)