Eigenvalues and eigenvectors of k-blocks matrix I'm trying to find the eigenvalues and eigenvectors of the following $n\times n$ matrix, with $k$ blocks.
\begin{gather*}
X = \left( \begin{array}{cc}
A & B & \cdots &  \\
B & A & B & \cdots \\
B & B & A & B & \cdots \\
\cdots\end{array} \right) \\
A = \left( \begin{array}{cc}
a & \cdots   \\
a & \cdots \\
\cdots\end{array} \right) \in R^{\frac{n}{k} \times \frac{n}{k}} \\
B = \left( \begin{array}{cc}
b & \cdots   \\
b & \cdots \\
\cdots\end{array} \right) \in R^{\frac{n}{k} \times \frac{n}{k}}.
\end{gather*}
I know that the first eigenvector is the constant $1$ vector. 
So the first eigenvalue is: $X1 = \left(\frac{n}{k}a+(k-1)\frac{n}{k} b\right)1$.
How do I find all of the rest of the eigenvalues and eigenvectors?
 A: $X$ is simply a tensor product $C\otimes D$ where $C$ is the matrix with all diagonal entries $a$ and non-diagonal entries $b$ and where each entry in $D$ is $1$.
If $R$, $S$ are diagonalizable, then the eigenvalues of $R\otimes S$ are simply the products $\lambda\mu$ where $\lambda$ is an eigenvalue of $R$ and $\mu$ is an eigenvalue of $S$. The eigenvectors of $R\otimes S$ are simply the values $x\otimes y$ where $x$ is an eigenvector of $R$ and $y$ is an eigenvector of $S$.
The eigenvalues of $D$ are $0$, $n/k$ where $n/k$ corresponds to the eigenvector $[1,\dotsc,1]^\perp$ and the eigenvectors of $0$ are the vectors $[x_1,\dotsc,x_{n/k}]^T$ with $x_1+\dotsb+x_{n/k}=0$. The eigenvalues of $C$ can be found in a similar way.
A: The block matrix $X$ is a particular one. Set $\frac{n}{k}=m$; you may diagonalize $A$ or $B$ by the same unitary $U$, thus taking the diagonal block matrix $V$ with diagonal blocks $U$, the matrix $D=VXV^*$ will have its $m\times m$ blocks, diagonals with one non zero entry as the leading entry $ma$ or $mb$. Apply a perfect shuffle (that is a permutation $P$) to $D$ so $PDP^*=M\oplus 0$, where $M=mbJ_k+m(a-b)I_k$; ($J_k $ is the all one matrix of dimension $k$, $I_k$ the  identity matrix).
