Prove that $\operatorname{trace}(A) = 0$ if and only if $A^2 = 0$.

Question

Let $A\in M_{n \times n}$ such that rank of $A$ is $1$. Prove that $\operatorname{trace}(A) = 0$ if and only if $A^2 = 0$.

Answer 1

This is false. Consider $A = \pmatrix{1 & 0 \\ 0 & -1}$.

Edit after question edited: If $A\in M_{n\times n}$ is rank $1$ then $A=uv^T$ for some $u,v\in M_{n\times 1}$. We also know that $\operatorname{trace}(A) = u^Tv=u\cdot v$ (can you see why?). So the condition that this matrix is traceless is that $u^Tv=0=v^Tu$. Then $$A^2 = (uv^T)(uv^T) = u(v^Tu)v^T = 0$$ by associativity of matrix multiplication. $\ \ \ \square$