Nilpotent endomorphism

Definition

Let $F$ be a field, let $V$ be a vector space over $F$, and let $T \colon V \to V$ be an endomorphism. The map $T$ is nilpotent when $T^k = 0$ for some integer $k \ge 1$. The least such $k$ is the index of $T$.

Remarks

If $\dim V = n$ is finite and $T$ is nilpotent, then $T^n = 0$. In finite dimensions, $T$ is nilpotent if and only if its characteristic polynomial is $t^n$. Over an algebraically closed field, this means that $0$ is its only eigenvalue. A matrix is nilpotent when its map is nilpotent, see nilpotent matrix. The derivative on the space of polynomials of degree less than $n$ is nilpotent.

Explanations

L1 human

A nilpotent map destroys everything after a few steps. Each application pushes vectors further towards zero, until they vanish. The shift $e_1 \mapsto 0$, $e_2 \mapsto e_1$, $e_3 \mapsto e_2$ in $F^3$ is nilpotent of index $3$.