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

Statement

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

If $V = 0$ , then

$∥ T ∥ = 0.$

If $V \neq 0$ and $s_{1}$ is the largest singular value of $T$ , then

$∥ T ∥ = s_{1} .$

In the nonzero case, the maximum in The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| is attained at a right-singular vector for $s_{1}$ .

Facts & Assumptions

Given: A linear map $T : V \to W$ between finite-dimensional real or complex inner product spaces.

[L1]

There is a singular value decomposition $T v = \sum_{j = 1}^{r} s_{j} ⟨ v, e_{j} ⟩ f_{j}$ with $s_{1} \geq \dots \geq s_{r} > 0$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct

1.1

If $V = 0$ , then The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| gives $∥ T ∥ = 0$ . Assume now that $V \neq 0$ , and choose an SVD from [L1]: $T v = \sum_{j = 1}^{r} σ_{j} ⟨ v, e_{j} ⟩ f_{j}, σ_{1} \geq \dots \geq σ_{r} > 0.$ Extend this list by zeros, setting $σ_{r + 1} = \dots = σ_{n} = 0$ ; then the displayed formula gives $T e_{j} = σ_{j} f_{j}$ for every $j$ , so $σ_{1}$ is the largest singular value of $T$ . Writing a unit vector as $v = \sum_{j} c_{j} e_{j}$ , one gets $∥ T v ∥^{2} = \sum_{j = 1}^{n} σ_{j}^{2} ∣ c_{j} ∣^{2} \leq σ_{1}^{2} \sum_{j} ∣ c_{j} ∣^{2} = σ_{1}^{2},$ so every unit vector satisfies $∥ T v ∥ \leq σ_{1}$ .

L1algebra

2.1

For the right-singular vector $e_{1}$ , step 1.1 gives $∥ T e_{1} ∥ = ∥ σ_{1} f_{1} ∥ = σ_{1}$ . Therefore the maximum defining $∥ T ∥$ equals the largest singular value $σ_{1}$ , and it is attained at $e_{1}$ when $V \neq 0$ .

L1step 1.1∎

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Used by

Dependency tree · two levels

7 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

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