CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Every idempotent endomorphism is diagonalisable and is projection onto its image along its kernel

Statement

If an endomorphism $T : V \to V$ of a finite-dimensional vector space satisfies $T^{2} = T$ , then it is diagonalisable and

$V = im T \oplus \ker T .$

Under this decomposition, $T$ is projection onto $im T$ along $\ker T$ .

Facts & Assumptions

Given: An idempotent endomorphism $T$ , so $T^{2} = T$ .

[L1]

For an endomorphism of a finite-dimensional space, a polynomial annihilates $T$ exactly when it is divisible by $μ_{T}$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L2]

For an endomorphism of a finite-dimensional space, having a product of distinct linear factors as minimal polynomial is equivalent to diagonalisability (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

[L3]

For two subspaces, a direct sum is a spanning sum with zero intersection (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

[L4]

The polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Proof

technique · direct

1.1L1L2L4algebra

The identity $T^{2} - T = 0$ says $x (x - 1)$ annihilates $T$ , so [L1] gives $μ_{T} ∣ x (x - 1)$ . By unique factorisation [L4], this monic divisor is a product of a subset of the two distinct irreducibles $x$ and $x - 1$ ; [L2] therefore makes $T$ diagonalisable.

1.2givenalgebra

Every $v \in V$ has $v = T v + (v - T v)$ , where $T v \in im T$ and $T (v - T v) = T v - T^{2} v = 0$ . Hence the image and kernel span $V$ .

2.1step 1.2L3algebra∎

If $w \in im T \cap \ker T$ , write $w = T u$ and compute $w = T u = T^{2} u = T w = 0$ . Thus [L3] and step 1.2 give the direct sum, and $T$ is identity on its image and zero on its kernel. The cases $T = 0$ and $T = I$ are included.

Depends on

Used by

Dependency tree · two levels

25 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

Keith Conrad, The Minimal Polynomial and Some Applications, Corollary 4.14 (standard reference, not scraped)