Let $x=\begin{bmatrix}3\cr4\end{bmatrix}$ and $A=\begin{bmatrix}0&x^T\cr x&0\end{bmatrix}$ is A diagonizable?

Question

I had a problem: let $x=\begin{bmatrix}3\cr4\end{bmatrix}$ and $A=\begin{bmatrix}0&x^T\cr x&0\end{bmatrix}$ is A diagonizable?

But when I plug in the matrix x and its transpose into A the dimensions don't work out and we have empty slots where there should be elements and I was wondering if I'm just tripping or if this problem is unsolvable?

Answer 1

Presumably the bottom right $0$ is actually a $2 \times 2$ matrix of $0$, i.e. $$A = \left[\matrix{0 & 3 & 4\cr 3 & 0 & 0\cr 4 & 0 & 0\cr} \right]$$