Given a non-null $3 \times 3$ matrix $A$ such that $A^2 = 0$, then prove that there exists a unitary matrix $U$ such that $$U^H A U = \begin{bmatrix} 0 & 0 & u \\ 0 & 0 & v \\ 0 & 0 & 0\end{bmatrix}$$
I tried taking a generic $A$ as: $$A=[v_1\:v_2\:av_1+bv_2]$$ where $v_1,v_2,av_1+bv_2$ are linearly dependent column vectors of $A$ since $Det(A)=0$. Now let the unitary matrix be: $$U=[u_1\:u_2\:u_3]$$ But the matrix multiplication is too heavy.
the result trivially holds when $A=\mathbf 0$ so we can assume WLOG that $A\neq \mathbf 0$.
1.) By Sylvester's Rank Inequality we have
$\text{rank}\big(A\big)+\text{rank}\big(A\big)\leq \text{rank}\big(A^2\big)+n = 0 +3\implies \text{rank}\big(A\big) = 1$
2.) By rank-nullity $\dim \ker \big(A\big) = 3-1=2$. So select 2 linearly independent vectors $\mathbf v_1, \mathbf v_2$ from the kernel. Rescale so that $\mathbf v_1$ has length (2 norm) of 1 and call it $\mathbf u_1$. Now run Gram Schmidt on $\mathbf v_2$ to get a second orthogonal vector that is re-scaled to have length one and call it $\mathbf u_2$.
3.) now find a 3rd basis vector, $\mathbf v_3$, that is linearly independent from the other $\mathbf v_k$'s. There are many ways to do this. An easy way is to note that at least one standard basis vector $\mathbf e_k$ must work. (Why? a simple dimension argument will do it.) Run Gram Schmidt on $\mathbf v_3$ as before to get $\mathbf u_3$.
Alternatively: consider the matrix $B:=\begin{bmatrix}\mathbf u_1^*\\\mathbf u_2^* \end{bmatrix}$
$2=\text{row rank}\big(B\big)=\text{col rank}\big(B\big)\implies \dim \ker\big( B\big) = 1$. So select a non-zero vector in that kernel and rescale so that it has length one, and call it $\mathbf u_3$.
4.)
$U := \bigg[\begin{array}{c|c|c} \mathbf u_1 & \mathbf u_2 & \mathbf u_{3} \end{array}\bigg]$
where $U$ is unitary. Then
$AU = U\begin{bmatrix}0 &0 & x_1\\ 0 &0 & x_2\\0 &0 & x_3 \end{bmatrix}= U\begin{bmatrix}0 &0 & x_1\\ 0 &0 & x_2\\0 &0 & 0 \end{bmatrix}= UR$
where we know $x_3=0$ because $A$ is nilpotent and hence has $n=3$ eigenvalues equal to zero (i.e. $A=URU^*$ and the eigenvalues of a triangular matrix are on the diagonal).