CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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An endomorphism and its adjoint have the same singular values

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional real or complex inner product space. Then $T$ and $T^{*}$ have the same singular values.

Facts & Assumptions

Given: An endomorphism $T : V \to V$ of a finite-dimensional real or complex inner product space.

[L1]

The singular value decomposition of $T$ has the form $T v = \sum_{j = 1}^{n} s_{j} ⟨ v, e_{j} ⟩ f_{j}$ for orthonormal bases $(e_{j})$ and $(f_{j})$ and singular values $s_{j} \geq 0$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct

1.1

By [L1], define $S : V \to V$ by $S w = \sum_{j = 1}^{n} s_{j} ⟨ w, f_{j} ⟩ e_{j}$ . For all $v, w \in V$ , orthonormality gives $⟨ T v, w ⟩ = \sum_{j} s_{j} ⟨ v, e_{j} ⟩ ⟨ f_{j}, w ⟩ = ⟨ v, S w ⟩$ , so $S = T^{*}$ .

L1algebra

2.1

The formula in step 1.1 is itself a singular value decomposition of $T^{*}$ , with the same diagonal coefficients $s_{j}$ . Therefore $T^{*}$ has the same singular values as $T$ .

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)