TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Characterisations of a nilpotent endomorphism

Statement

Let $N$ be an endomorphism of a nonzero $n$ -dimensional vector space over $F$ . The following are equivalent:

$N$ is nilpotent;
$μ_{N} = x^{r}$ for some $1 \leq r \leq n$ ;
$χ_{N} = x^{n}$ ;
some ordered basis gives $N$ a strictly upper-triangular matrix.

In that case $r$ is the nilpotency index.

The nonzero hypothesis is needed for condition 2. On the zero space the unique endomorphism $N = 0$ is nilpotent with nilpotency index $1$ , and $μ_{N} = χ_{N} = 1 = x^{0}$ while the empty matrix is strictly upper triangular; so conditions 1, 3 and 4 hold there, but condition 2 fails, since no integer $r$ satisfies $1 \leq r \leq 0$ . The exponent $0$ of $μ_{N}$ is then not the nilpotency index $1$ .

Facts & Assumptions

Given: An endomorphism $N$ of a finite-dimensional $F$ -vector space $V$ , with $n = \dim V$ .

[L1]

A polynomial $p$ annihilates an endomorphism exactly when its minimal polynomial divides $p$ ; on the zero space the minimal polynomial is $1$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L2]

The minimal and characteristic polynomials have exactly the same monic irreducible factors (The minimal and characteristic polynomials have exactly the same monic irreducible factors).

[L3]

An endomorphism is triangularisable exactly when its characteristic polynomial splits ( $T$ is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits).

[L4]

The characteristic polynomial is monic of degree $n$ , and it is $1$ when $n = 0$ ( $χ_{A} (x)$ 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 $χ_{0 \times 0} = 1$ ).

[L5]

Every endomorphism satisfies its characteristic polynomial (Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, $χ_{T} (T) = 0$ ).

Proof

technique · direct

1.1

Suppose $n > 0$ . By [L1], $N^{k} = 0$ for some positive $k$ exactly when $μ_{N} ∣ x^{k}$ , which is exactly when $μ_{N} = x^{r}$ for some positive $r$ ; [L5] then bounds the least such $r$ by $n$ once $χ_{N} = x^{n}$ .

L1L5

1.2

If $μ_{N} = x^{r}$ , [L2] says that $x$ is the only irreducible factor of $χ_{N}$ , and [L4] forces $χ_{N} = x^{n}$ ; conversely $χ_{N} = x^{n}$ and [L5] give $N^{n} = 0$ .

L2L4L5

2.1

If $N$ is nilpotent, step 1.2 makes $χ_{N}$ split, so [L3] gives an upper-triangular matrix; its diagonal entries are roots of $χ_{N} = x^{n}$ , hence are all zero, making it strictly upper triangular. Conversely, the $n$ th power of a strictly upper-triangular $n \times n$ matrix is zero.

step 1.2L3algebra

3.1

For $V = 0$ , [L1] and [L4] give $μ_{N} = χ_{N} = 1 = x^{0}$ , the unique empty matrix is strictly upper triangular, and the convention in Nilpotent endomorphisms and their nilpotency index gives index $1$ , so conditions 1, 3 and 4 hold; condition 2 asks for $r$ with $1 \leq r \leq 0$ and no such integer exists, and the exponent $0$ of $μ_{N}$ differs from the index $1$ , which is why the equivalence is stated for $n > 0$ . Together with steps 1.1-2.1 this proves every asserted case.

step 1.1step 1.2step 2.1L1L4∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Results 8.17-8.18 (standard reference, not scraped)