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Here's what I know how to do. $(B^2)^2 = B^2 B^2$ $ B^2 B^2 = B^2$ $B^2$ commutes with itself. $B^2$ is diagonal (if it's two times a $2 \times 2$ matrix, $B$ is definitely diagonal). $B^2$ is hermitian. $B^2$ is real. $B^2$ is symmetric. Now, somehow, the matrix $B^2$ should magically appear diagonally. It's probably somehow related to the fact that it's diagonal (or at least regular), so I suspect that I'm missing some special property of $B$ that I can use. So, what is this property? Thanks in advance. A: I would start the proof off like this: $$B^3 = B^2\implies\begin{bmatrix}b_{11}& b_{12}\\b_{21} & b_{22}\end{bmatrix}\begin{bmatrix}b_{11} & b_{21}\\b_{12} & b_{22}\end{bmatrix} = \begin{bmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{bmatrix}$$ Note that since this is a $2\times 2$ matrix, $b_{12} = b_{21}$. Also, since $B$ is invertible, we have that $b_{11}b_{22} - b_{21}^2 = 1$. 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