How to diagonalize a special block two-by-two matrix

Question

I have a special block two-by-two matrix as \begin{equation} \mathcal{A}=\begin{bmatrix} & -I_n\\ A & \end{bmatrix}, \end{equation} where $I_n$ is an identity matrix of order $n$ and $A\in\mathbb{R}^{n\times n}$ is a symmetric positive definite matrix with the eigenvalue decomposition $A = V_AD_AV^{-1}_A$. Can I prove that the matrix $\mathcal{A}$ is also diagonalizable? If yes, can I write the explicit form of its eigenvalue decomposition?

Answer 1

As $A$ and $-I_n$ commute, you can just diagonalize this in the usual way, treating $A$ and $-I_n$ as numbers. This will seem a little unusual as you will have to take the square root of $A$, but this is not a problem as $A$ is positive definite. So, in the first step you end up with the block diagonal matrix $\tilde{D} := \begin{pmatrix} -i\sqrt{A} & 0 \\ 0 & i\sqrt{A} \end{pmatrix}$ and the transformation matrix $\tilde{V} := \begin{pmatrix} -i\sqrt{A^{-1}} & i\sqrt{A^{-1}} \\ I_n & I_n \end{pmatrix}$. The two blocks of $\tilde{D}$ can now be diagonalized as usual, using $\sqrt{A} = V_A\sqrt{D_A}V_A^{-1}$.