Given diagonalizable $n \times n$ matrix $A$ ($A = PDP^{-1}$, where D is diagonal matrix). How can I prove that
$$ \left[\begin{matrix} A, & -A \\ -A, & A \\ \end{matrix} \right] $$
is also diagonalizable?
I tried constructing a diagonalization for the block matrix, but with no luck. $$ \left[\begin{matrix} A & -A \\ -A & A \\ \end{matrix} \right] = \left[\begin{matrix} P_1 & P_2 \\ P_3 & P_4 \\ \end{matrix} \right] \left[\begin{matrix} D_1 & 0\\ 0 & D_2 \\ \end{matrix} \right] \left[\begin{matrix} P_1 & P_2 \\ P_3 & P_4 \\ \end{matrix} \right]^{-1} $$
This becomes particularly easy if you know about the Kronecker product since your matrix is simply $$ M = \pmatrix{1 & -1\\-1 & 1} \otimes A. $$
Another quick approach is to start by considering the similar matrix $$ \pmatrix{I & I\\I & -I}^{-1}\pmatrix{A & -A\\-A & A}\pmatrix{I & I\\I & -I} = \pmatrix{0 & 0\\0 & 2A} , $$ where $I$ denotes the identity matrix, and we note that $$ \pmatrix{I & I\\I & -I}^{-1} = \frac 12 \pmatrix{I & I\\I & -I}. $$
I think that the nicest direct approach, however, is as follows. Suppose that $z = (z_1,z_2)$ (where $z,z_1,z_2$ are column vectors with $z_1,z_2 \in \Bbb R^n$) is an eigenvector of $M$. That is, $$ Mz = \lambda z \implies \pmatrix{A & -A\\-A & A} \pmatrix{z_1\\z_2} = \lambda \pmatrix{z_1 \\ z_2} \implies \begin{cases} (A - \lambda I)z_1 = Az_2\\ (A - \lambda I)z_2 = Az_1 \end{cases}\\ \implies \begin{cases} (A - \lambda I)(z_1 + z_2) = A(z_1 + z_2)\\ (A - \lambda I)(z_1 - z_2) = -A(z_1 - z_2) \end{cases} \\\implies \begin{cases} \lambda(z_1 + z_2) = 0,\\ (2A - \lambda I)(z_1 - z_2) = 0. \end{cases} $$ To go from the first system to the second system: add the equations to produce the first new equation and subtract the equations to produce the second new equation. To go from the second system to the third, just simplify.
With the above analysis, we can conclude that if $A$ has a basis of eigenvectors, then $M$ also has a basis of eigenvectors.