Let $A, B \in {\Bbb R}^{2 \times 2}$ be two self-adjoint matrices. I am interested in the following block matrix
$$ M = \begin{bmatrix} A & B & B & \dots & B \\ B & A & B & \dots & B \\ B & B & A & \dots & B \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ B & B & B & \dots & A \end{bmatrix}$$
of size $2m \times 2m$. I am curious if there is a way to diagonalize this matrix explicitly in the sense that one can reduce it to the matrices $A$ and $B$?
A variant of Willie Wong's answer that gives an explicit eigendecomposition.
Let $C = A - B$, and $E$ be the matrix of all ones. Your matrix is $M = I\otimes C + E \otimes B$, where $\otimes$ is the Kronecker product. Let $E = QDQ^*$ be an eigendecomposition. Then, $$ M = (Q\otimes I) (I\otimes C + D \otimes B) (Q\otimes I)^*. $$ Since $D = \operatorname{diag}(n,0,0,\dots,0)$, the matrix in the middle is the block diagonal matrix $$ (Q\otimes I)^*M(Q\otimes I) = I\otimes C + D \otimes B = \operatorname{diag}(C+nB, C, C, \dots, C). $$ If $C+nB = Q_1 D_1 Q_1^*$ and $C = Q_2 D_2 Q_2^*$, then you can reduce each diagonal block the matrix to diagonal form with $$ \begin{multline} \operatorname{diag}(Q_1,Q_2,Q_2,\dots,Q_2)^*(Q\otimes I)^*M(Q\otimes I)\operatorname{diag}(Q_1,Q_2,Q_2,\dots,Q_2) \\= \operatorname{diag}(D_1,D_2,D_2,\dots,D_2). \end{multline} $$
Write $C = A - B - \lambda I_2$. You want to compute
$$ \begin{vmatrix} C + B & B & \ldots \\ B & C + B & \ldots \\ \vdots & \vdots & \ddots \end{vmatrix} $$
enlarge the matrix by two rows and two columns, to get
$$ = \begin{vmatrix} I_2 & B & B & B & \ldots \\ 0 & C+ B & B & B & \ldots \\ 0 & B & C+B & B & \ldots \\ \vdots & \vdots & \vdots & & \ddots \end{vmatrix} $$
subtract the first two rows from each pairs of the rows below, we get
$$ = \begin{vmatrix} I_2 & B & B & B & \ldots \\ -I_2 & C & 0 & 0 & \ldots \\ -I_2 & 0 & C & 0 & \ldots \\ \vdots & \vdots & \vdots & & \ddots \end{vmatrix} $$
This block matrix satisfies the hypotheses of the main theorem in this paper (as all entries that are not the first row commute with each other). So the determinant evaluates as
$$ \det( C^m + m B C^{m-1} ) = \det( C + m B) \det(C)^{m-1} $$
Converting back to the original notation, you see that the eigenvalues are
(assuming I didn't make a sign error somewhere).
For eigenvectors:
Doing this brute force by using Mathematica for low values of $m$, one easily finds the pattern for the eigenvalues, which is perhaps not what you were hoping for.
Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ and $B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$.
Then, the eigenvalues are given by $$ \lambda = \frac{1}{2} ( a_{11} + a_{22} - b_{11} - b_{22} \pm \sqrt{ (a_{11} + a_{22} - b_{11} - b_{22})^2 + 4 (a_{12} a_{21} - a_{11} a_{22} + a_{22} b_{11} - a_{21} b_{12} - a_{12} b_{21} + b_{12} b_{21} + a_{11} b_{22} - b_{11} b_{22}) } ) $$ which are both $m-1$ fold degenerate, and $$ \lambda = \frac{1}{2} ( a_{11} + a_{22} - (m-1)(b_{11} + b_{22}) \pm \sqrt{ (a_{11} + a_{22} - b_{11} - b_{22})^2 + 4 (a_{12} a_{21} - a_{11} a_{22} - (m-1) (a_{11} b_{22} + a_{22} b_{11} - a_{21} b_{12} - a_{12} b_{21}) + (m-1)^2 (b_{12} b_{21} - b_{11} b_{22})) } ) $$
Given the rather simple structure of the eigenvalues, it should not be too hard to prove this. Doing a Fourier transform should do the trick.
