Eigenvalues and eigenvectors of k-blocks matrix

Question

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?

Answer 1

$X$ 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.

Answer 2

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).