Prove that $A$ is diagonalizable if and only if $\mbox{tr} A\neq 0$.

$A$ is an $n\times n$ matrix over $\mathbb{C}$, and $\mbox{rk} A=1$.

If $p(t)$ is the characteristic polynomial of $A$, I know that $a(n-1)\neq0$ because $\mbox{tr} A = (-1)^{n+1}a(n-1).$
I also know that $\dim\ker(A-0\cdot I)=\dim\ker A=n-\mbox{rk} A=n-1$ (so the geometric multiplicity of $t=0$ as an eigenvalue is $n-1$).
Though I don't know how to continue from here (on both directions).

Any suggestions?
Thanks