TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Every endomorphism has a polar decomposition T = SU with U non-negative and S an isometry on the orthogonal complement of ker T, and S is unique exactly when T is invertible

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional real or complex inner product space. Then there exist a non-negative endomorphism $U$ and a linear isometry $S : V \to V$ such that

$T = S U and U = \sqrt{T^{*} T} .$

Equivalently, $S$ is an isometry on $(\ker T)^{⊥} = im U$ . The factor $S$ is unique if and only if $T$ is invertible.

Facts & Assumptions

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

[L1]

There are orthonormal bases $(e_{j})$ and $(f_{j})$ and singular values $s_{1} \geq \dots \geq s_{r} > 0 = s_{r + 1} = \dots = s_{n}$ such that $T e_{j} = s_{j} f_{j}$ for all $j$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

A linear map that sends an orthonormal basis to an orthonormal basis is a linear isometry (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

[L3]

A non-negative operator has a unique non-negative square root (A non-negative operator has a unique non-negative square root).

Proof

technique · direct

1.1

By [L1], define $U e_{j} : = s_{j} e_{j}$ and $S e_{j} : = f_{j}$ for every $j$ . Then $S$ is a linear isometry by [L2], $S U e_{j} = s_{j} f_{j} = T e_{j}$ for every basis vector, hence $T = S U$ , and $U^{2} e_{j} = s_{j}^{2} e_{j} = T^{*} T e_{j}$ . Thus $U^{2} = T^{*} T$ , and because $U$ is self-adjoint with nonnegative eigenvalues in the basis $(e_{j})$ , [L3] identifies $U$ with the unique non-negative square root $\sqrt{T^{*} T}$ .

L1L2L3algebra

2.1

If $T$ is invertible, then all singular values are positive, so $U$ is invertible. Any other factorisation $T = S^{'} U$ with the same non-negative factor $U = \sqrt{T^{*} T}$ satisfies $S^{'} = T U^{- 1} = S$ , so $S$ is unique.

step 1.1algebra

3.1

If $T$ is not invertible, then $r < n$ . Define another isometry $S^{'}$ by $S^{'} e_{j} = f_{j}$ for $j \neq r + 1$ and $S^{'} e_{r + 1} = - f_{r + 1}$ . Then $S^{'} \neq S$ , but $S^{'} U e_{j} = S U e_{j} = T e_{j}$ for $j \leq r$ and $S^{'} U e_{j} = 0 = T e_{j}$ for $j > r$ because $s_{j} = 0$ . Hence $S^{'} U = T$ , so uniqueness fails.

step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

11 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)