$J_n(0)$ Nilpotent?

Question

Given the matrix $M = J_n(0)$, i.e the matrix M is just a jordan block with all eigenvalue $0$, how would one show this matrix is nilpotent? I can visualize that matrix $M$ takes the form $$ M = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\0 & 0 & 1 & \cdots & 0 \\\vdots & \vdots & \ddots & \ddots & \vdots \\0 & 0 & \cdots & & \\0 & 0 & \cdots & 0 & 0\end{bmatrix},$$ I can see that eventurally $M^k$ will termintate, for some $k \in \mathbb{Z}^+$.

Answer 1

Let $e_i$ be the $i$'th standard basis vector, and define $e_0 = 0$. Then $M e_i = e_{i-1},$ for $i =1,..., n$ and clearly $M e_0 = 0.$ Thus $M^n e_i = 0$ for all $i.$

Answer 2

$M = J_n(0)$ has only one eigenvalue: $ \lambda =0$. Hence the characteristic polynomial $p$ of $M$ is given by

$$p(x)=x^n.$$

By Cayley - Hamilton it follows that

$$0=p(M)=M^n.$$