Show that if an $n \times n$ matrix, $A$, is nilpotent, then $\text{rank}(A) < n$

Question

A matrix is $A$ is nilpotent if $A^k$ is the $0$ matrix for some $k$.

Show that if an $n \times n$ matrix $A$ is nilpotent, then $\text{rank}(A) < n$.

Anyone got any ways to get started on this problem?

Answer 1

If $A$ is nilpotent, then $A^k=0$ for some $k$. Then $\det(A^k) = 0$, but $\det(A^k)=\det(A)^k$. So what can you say about $\det(A)$, and what does that tell you about the rank of $A$?

Answer 2

Any nilpotent matrix is similar to an upper triangular matrix with zeros on the diagonal, by a theorem. Actually it's a special case of the Jordan canonical form.

But such a matrix has $\rm{rank}\lt n$.