LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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$χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$

Statement

For $A \in M_{n} (F)$ , the polynomial $χ_{A} (x)$ is monic of degree $n$ . If $n \geq 1$ , the coefficient of $x^{n - 1}$ is $- tr (A)$ and the constant coefficient is $(- 1)^{n} \det (A)$ . For $n = 0$ , $χ_{A} (x) = 1$ .

Facts & Assumptions

Given: A matrix $A = (a_{i j}) \in M_{n} (F)$ .

[L1]

For $n \geq 1$ , $χ_{A} (x) = \det (x I_{n} - A)$ ; for $n = 0$ , it is $1$ (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

[L2]

For $n \geq 1$ , $\det (C) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} c_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

The trace of $A$ is $\sum_{i < n} a_{i i}$ (The trace $tr (A)$ as the sum of the diagonal entries).

[L4]

The degree of a nonzero polynomial is the largest index of a nonzero coefficient, its leading coefficient is the coefficient at that index, and it is monic when that leading coefficient is $1$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

Proof

technique · direct

1.1

If $n = 0$ , [L1] gives $χ_{A} = 1$ , which is monic of degree $0$ ; the additional coefficient assertions are restricted to $n \geq 1$ .

L1L4

1.2

Assume $n \geq 1$ . In [L2] for $x I_{n} - A$ , the identity permutation contributes $\prod_{i < n} (x - a_{i i})$ . Every nonidentity permutation moves at least two indices, so its term contains at most $n - 2$ diagonal factors and has degree at most $n - 2$ .

L1L2algebra

1.3

Evaluating at $x = 0$ gives the constant coefficient $\det (- A) = (- 1)^{n} \det (A)$ directly from [L2].

L1L2algebra

2.1

Hence the coefficient of $x^{n}$ is $1$ , the coefficient of $x^{n - 1}$ comes only from the identity term and is $- \sum_{i < n} a_{i i} = - tr (A)$ , and there are no terms of degree above $n$ . Thus $χ_{A}$ is monic of degree $n$ .

step 1.2L3L4algebra

3.1

Steps 1.1–2.1 establish every asserted dimension and coefficient case.

step 1.1step 2.1step 1.3∎

Depends on

Used by

Dependency tree · two levels

17 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

H. Pinkham, Linear Algebra, §12.1 (standard reference, not scraped)
M. Khovanov, Linear Algebra II notes, §6 (standard reference, not scraped)

χA(x)χA​(x) is monic of degree nn; for n≥1n≥1 its xn−1xn−1 coefficient is −tr⁡(A)−tr(A) and its constant coefficient is (−1)ndet⁡(A)(−1)ndet(A), while χ0×0=1χ0×0​=1