Prove that a matrix $A$ is nilpotent iff for all $v$ there is $k$ such that $A^k v=0$

Question

For nilpotent matrices, I was given two equivalent definitions. Let $A \in End(V)$.

1. $\forall v \in V$ there exists $k \geq0$ such that $A^k v = 0$.

2. there exists $k \geq 0$ such that $\forall v \in V$, $A^kv=0$.

They seem to be equivalent but i don't know how to show it formally.

Answer 1

Let $e_1,e_2,..,e_n$ be basis. Under 1) there exists $k_1,k_2,..,k_n$ such that $A^{k_i}e_i=0$ for each $i$. It follows that $A^{k}e_i=0$ for each $i$ if $k$ exceeds each of the numbers $k_1,k_2,...,k_n$. Since $e_1,e_2,..,e_n$ is basis it follows that $A^{k}v=0$ for all $v$ for such $k$.

Answer 2

$1\Rightarrow2$

You have $A=0$ iff $Av=0$ for all $v$ in a basis. Then take an orthonormal basis $\{e_1,...,e_n\}$. For each $e_j$ there is by hypothesis some $k_j\in\mathbb N$ such that $A^{k_j}e_j=0$. Thus if you take the biggest of these $k_j$ you obtain some $k$ such that $A^k e_j=0$ for all $j=1,...,n$, and therefore $A=0$.

$1\Leftarrow2$

If $A^k=0$ then $A^k v=0$ for all $v$, and therefore the statement is true taking the same $k\in\mathbb N$ for each vector $v$.