False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16

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FALSE: Every idempotent endomorphism of an inner product space is an orthogonal projection

Statement

False claim. Every idempotent endomorphism of an inner product space is an orthogonal projection.

Facts & Assumptions

Given: In standard $R^{2}$ , the matrix $P = (\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array})$ .

[L1]

An endomorphism is an orthogonal projection exactly when it is both idempotent and self-adjoint (An endomorphism is an orthogonal projection exactly when it is idempotent and self-adjoint).

[L2]

In an orthonormal real basis, the adjoint matrix is the transpose (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix).

Refutation

technique · counterexample

1.1algebra

Direct multiplication gives $P^{2} = P$ , so $P$ is idempotent.

1.2L1L2

But $P^{T} = (\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array}) \neq P$ , so [L2] shows that $P$ is not self-adjoint. Hence [L1] shows it is not an orthogonal projection.

2.1algebra∎

Concretely, $im P = span (1, 0)$ and $\ker P = span (- 1, 1)$ , whose generators have dot product $- 1$ , so this is an oblique projection.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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, 4th ed., §§6C and 7A (standard reference, not scraped)