$T$ is nilpotent if and only if $[T]_\beta$ is nilpotent.

Question

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\beta$ be an ordered basis for $V$. Prove that $T$ is nilpotent if and only if $[T]_\beta$ is nilpotent, where $[T]_\beta$ denotes the matrix of $T$ w.r.t ordered basis $\beta$ .

Let $p$ be the degree of nilpotency of $T$. So $T^p = T_0$ and $[T^p]_\beta = 0$. It will be enough if we show $([T]_\beta)^p=0$. For that we have to show that $$[T^p]_\beta = ([T]_\beta)^p.$$

Facing problem to show the above result.

Answer 1

More generally, if $S$ and $T$ are any two operators on V, $\;\;\big[S\circ T\big]_{\beta}=\big[S\big]_{\beta}\big[T\big]_{\beta}$ since

$\big[S\circ T\big]_{\beta}\big[v\big]_{\beta}=\big[(S\circ T)(v)\big]_{\beta}=\big[S(T(v))\big]_{\beta}=\big[S\big]_{\beta}\big[T(v)\big]_{\beta}=\big[S\big]_{\beta}\big(\big[T\big]_{\beta}\big[v\big]_{\beta}\big)=\big(\big[S\big]_{\beta}\big[T\big]_{\beta}\big)\big[v\big]_{\beta}$

for any $v\in V$, so using induction gives $\big[T^p\big]_{\beta}=\big(\big[T\big]_{\beta}\big)^p$.