How to prove that $A$ (a $n\times n$ matrix) is diagonalizable, given that $rank(A) = 1$ and $A^2 != 0$?

Question

So I have proved that $0$ is an eigenvalue for the above matrix and that the geometric multiplicity of $0$ is $n-1$. I know if I can find one more eigenvector for some other eigenvalue, I will be able to prove this. But I don't know how to find this other eigenvector.

Answer 1

If $\operatorname{rank}(A)=1$ then $A=vw^T$ for some non-zero vectors $v,w$. Then $Av=vw^Tv=\langle w,v\rangle v$.

Answer 2

Hint: If $x$ is in the column space of $A$, then $Ax$ must also be in the column space of $A$.