LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A Hermitian idempotent has trace equal to its rank

Statement

If $P \in C^{n \times n}$ is Hermitian and $P^{2} = P$ , then $tr P = rank P$ , where rank is the dimension of the image (Rank and nullity of a linear map with finite-dimensional domain).

Facts & Assumptions

Given: A finite-dimensional complex inner-product space and a Hermitian idempotent endomorphism $P$ .

[F1]

Every normal endomorphism of a finite-dimensional complex inner-product space has an orthonormal eigenbasis (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

[F2]

The rank of an endomorphism is the dimension of its image (Rank and nullity of a linear map with finite-dimensional domain).

[F3]

Similar square matrices over a field have equal traces (Similar matrices have the same trace).

Proof

1.1givenF1choosealgebra

By [F1], choose an orthonormal eigenbasis for $P$ with eigenvalues $λ_{1}, \dots, λ_{n}$ . Idempotence gives $λ_{j}^{2} = λ_{j}$ , so each $λ_{j}$ is $0$ or $1$ .

2.1F2F3step 1.1algebra∎

Let $U$ be the unitary matrix whose columns are the eigenbasis from step 1.1, so $U^{*} P U = diag (λ_{1}, \dots, λ_{n})$ . Since $U$ is unitary, $U^{*} = U^{- 1}$ and [F3] gives $tr P = tr (U^{*} P U) = \sum_{j} λ_{j}$ , where the last equality is the diagonal definition of trace. The image is spanned exactly by the eigenvectors with eigenvalue $1$ , so by [F2] its dimension, and thus the rank, is the number of indices with $λ_{j} = 1$ . Therefore trace and rank agree.

Depends on

Used by

Cited to discharge well-definedness by Rank-r orthogonal projector domain.

Dependency tree · two levels

23 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

Alan Edelman, Tomás A. Arias, and Steven T. Smith, The Geometry of Algorithms with Orthogonality Constraints, SIAM Journal on Matrix Analysis and Applications 20 (1998), §§2.3.2 and 2.5 (standard reference, not scraped)
P.-A. Absil, R. Mahony, and R. Sepulchre, Optimization Algorithms on Matrix Manifolds, Chapter 3, Princeton University Press (2008), §3.4.4 (standard reference, not scraped)