TITLE: does a matrix like this exist?
QUESTION [7 upvotes]: Question:
Does a matrix $A \in M_{3 \times 3}(F)$ exist s.t. $A^4=
\begin{bmatrix} 0&0&1\\0&0&0\\0&0&0\end{bmatrix}$
What I thought:
I think it doesn't. How do you start a proof of such a thing (prefer hints at first).
Thx

REPLY [3 votes]: It's easy to compute $(A^4)^2=O$ (the null matrix).
I believe you already know the following result, but I'll prove it anyway, as it's useful in many situations.
If $\lambda$ is an eigenvalue of $A$ (in the complex numbers), then $\lambda^n$ is an eigenvalue of $A^n$: just do it by induction. The base step $n=1$ is obvious; then, if $v$ is an eigenvector for $\lambda$, you have, for $n>1$,
$$
A^nv=A(A^{n-1}v)=A(\lambda^{n-1}v)=\lambda^{n-1}Av=\lambda^nv.
$$
In our case this means that the only eigenvalue of $A$ is $0$. So the characteristic polynomial of $A$ has a very simple form. Apply Hamilton-Cayley, to conclude that $A^3=\dots$ and get a contradiction.