Eigenvalues and eigenvectors of a block matrix [A,B; B,A] Consider a block matrix of the type
$ M =  \begin{bmatrix}A & B \\B & A \end{bmatrix} $
where $A$ and $B$ are square matrices of equal dimensions. We know that
$\det(M) = \det(A-B) \det(A+B)$.
This property is very useful for computing the eigenvalues of $M$, because it implies that
$\det(M-\lambda I) = \det(A-B -\lambda I) \det(A+B -\lambda I)$,
(where $\lambda$ is a scalar and $I$ the identity matrix). Therefore, the eigenvalues of $M$ will be given by joining the sets of the eigenvalues of the matrices $A+B$ and $A-B$. This simplifies the problem because it reduces the dimensions of the matrices.
Is there a similar "trick" that we can exploit to simplify the calculations of the eigenvectors? That is, given a matrix $M$ in the block form shown above, is there a way to obtain its eigenvectors by calculating the eigenvectors of the smaller matrices $A$ and $B$ (or a linear combination of them) ?
 A: I have recently faced this problem. I first observed that eigenvectors seemed particular indeed. I finally noticed that, if A-B and A+B were diagonalizable (and their eigenvectors known) then eigenvectors of M could be deduced.
Assuming A+B and A-B are diagonalizable:
\begin{equation}P^{-1}_{A+B}(A+B)P_{A+B}=D_{A+B}\end{equation}
\begin{equation}P^{-1}_{A-B}(A-B)P_{A-B}=D_{A-B}\end{equation}
Let's construct these 2 new matrices:
\begin{equation}P_{M}=\begin{bmatrix}P_{A+B}&-P_{A-B}\\P_{A+B}&P_{A-B}\end{bmatrix}\end{equation}
\begin{equation}P^{-1}_{M}=1/2\begin{bmatrix}P^{-1}_{A+B}&P^{-1}_{A+B}\\-P^{-1}_{A-B}&P^{-1}_{A-B}\end{bmatrix}\end{equation}
We can check that:
\begin{equation}P_{M}P^{-1}_{M}=P^{-1}_{M}P_{M}=I\end{equation}
\begin{equation}P^{-1}_{M}MP_{M}=\begin{bmatrix}D_{A+B}&0\\0&D_{A-B}\end{bmatrix}\end{equation}
Within the scope of the hypotheses, eigenvectors of M can then be deduced with the knowledge of the ones of A-B and A+B. I don't know if this is generalizable when A-B and A+B are not simultaneously diagonalizable.
