Proof that Jordan form with eigenvalues of 0 is nilpotent

Question

I can see that if I have a nxn matrix Of Jordan form with eigenvalues of zero

$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}$

How do I prove that $A^n = 0$

Thank you

Answer 1

The linear map associated to such a matrix does the following to the standard basis $e_1,\dots,e_n$:

maps $e_n$ to $e_{n-1}$,

maps $e_{n-1}$ to $e_{n-2}$

...

maps $e_1$ to $0$.

Therefore, in $n$ steps all vectors of the basis are mapped to $0$. Hence $A^nX=0$ for all vector $X$, hence $A^n=0$.

Answer 2

We may obtain the result by matrix multiplication:

$$A^1 = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$ $$A^2 = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $$A^3 = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

Hence, $A$ is nilpotent in 3.