Any Nilpotent Matrix is not Diagonalizable I'm trying to go about the proof that any matrix that is nilpotent
(i.e. $\exists N \in\Bbb N. A^N = \mathbf{0}$) cannot be diagonalizable.
I believe that the best way to go about this is by showing
that a given eigenvalue's
geometric multiplicity is not the same as its algebraic multiplicity. However,
I am having some difficulty figuring out what aspect of nilpotency might
help me with this calculation.
I can see that if $A^N = \mathbf{0}$ for some $N \in\Bbb N,$ I think
it may be reasonable to prove that the only eigenvectors we get from $A$
are the $\mathbf{0}$ vector, although I am not entirely sure if this is the
appropriate way to prove the above statement, given that it doesn't
necessarily pertain to multiplicity but more of the mechanics of finding
eigenvectors and eigenvalues of a matrix. Any recommendations on this problem?

One line: Minimal polynomial divides $z^n=0$ so it must have repeated roots. 
*Result used: A complex matrix is diagonalizable iff its minimal polynomial has no repeated roots. If $p(A)=0$, minimal polynomial of $A$ divides $p(A)$. If a real matrix cannot be diagonalizable wrt to $\mathbb{C}$ then it surely cannot wrt to $\mathbb{R}$. 
