LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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The trace of an idempotent is its rank as a field scalar

Statement

If $P : V \to V$ is an idempotent endomorphism of a finite-dimensional space over a field $k$ , then $tr (P) = (\dim_{k} im P) \cdot 1_{k}$ . In positive characteristic this equality does not in general determine the integer rank from the trace.

Facts & Assumptions

Given: $P \in {End}_{k} (V)$ , $\dim_{k} V < \infty$ , and $P^{2} = P$ .

[F1]

$V = im P \oplus \ker P$ , with $P$ the projection onto its image (Every idempotent endomorphism is diagonalisable and is projection onto its image along its kernel).

[F2]

Trace is the matrix trace in any ordered basis, with trace zero on the zero space (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F3]

Every linearly independent subset of a subspace of a finite-dimensional vector space extends to a finite basis of that subspace, without a choice principle (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1

By F1, both $im P$ and $\ker P$ are subspaces of the finite-dimensional space $V$ . Apply F3 to each empty independent subset to obtain finite bases, and concatenate them in the displayed direct-sum order, putting the $r = \dim_{k} im P$ image basis vectors first. For $u = P (v)$ , $P (u) = P^{2} (v) = P (v) = u$ ; for $w \in \ker P$ , $P (w) = 0$ . Thus the matrix is $diag (I_{r}, 0)$ .

F1F3given

2.1

F2 permits this basis for computing trace. Summing the diagonal gives $tr (P) = \underset{r terms}{\underset{⏟}{1_{k} + \dots + 1_{k}}} = r \cdot 1_{k}$ . For $r = 0$ this is the empty sum $0$ ; when $V = 0$ it also agrees with F2. For $P = {id}_{V}$ , it gives $(\dim_{k} V) \cdot 1_{k}$ .

step 1.1F2

3.1

If $char (k) = p > 0$ , take $V = k^{p}$ . Its identity has rank $p$ and trace $p \cdot 1_{k} = 0$ , whereas its zero map has rank $0$ and trace $0$ . Therefore equal traces need not imply equal integer ranks, even among idempotents on the same space.

step 2.1algebra∎

Sources

Axler, 8.47–8.51, pp. 326–327, gives the trace convention and basis independence. Etingof et al., Theorem 4.5.1 proof, p. 68, uses projector trace over the complex numbers; the local proof retains arbitrary characteristic.

Depends on

Used by

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Sources

Etingof et al., Introduction to Representation Theory (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)