False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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FALSE: The singular values of an operator are the absolute values of its eigenvalues

Statement

The singular values of an operator are the absolute values of its eigenvalues.

Facts & Assumptions

Given: The nilpotent matrix $N = (\begin{array}{cc} 0 & 2 \\ 0 & 0 \end{array})$ on $R^{2}$ .

[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).

Refutation

technique · direct

1.1

The characteristic polynomial of $N$ is $x^{2}$ , so both eigenvalues are $0$ and their absolute values are $0, 0$ .

algebra

2.1

Direct multiplication gives $N^{T} N = diag (0, 4)$ , so the singular values are $2$ and $0$ . Because $2 \neq 0$ , the singular values are not the absolute values of the eigenvalues.

L1step 1.1algebra∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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)