We have matrix $C$ of the form:
$C =\begin{bmatrix} B_{1,1} A_{1,1} & B_{1,2} A_{1,2} & \dots & B_{1,K} A_{1,K} \\ B_{2,1} A_{2,1} & B_{2,2} A_{2,2} & \dots & B_{2,K} A_{2,K}\\ & & \ddots & \\ B_{K,1} A_{K,1} & B_{K,2} A_{K,2} & \dots & B_{K,K} A_{K,K} \end{bmatrix}$
where $B_{i,j}$ is a scalar, and $A_{i,j}$ is a matrix.
We also know that $B$ is a Markov (transition) matrix (the matrix consisted of $B_{i,j}$) , and each block $A_{i,j}$ is also Markov matrix.
I can numerically verify that eigenvalues of the $B$ matrix will be contained within the eigenvalues of $C$. (Eigenvalues of $B$ are also eigenvalues of $C$.) However, I can't mathematically prove this. Does anyone have any suggestion?
In the Kronecker product case $(C = B \otimes A)$, this is easy to see, since eigenvalues will be the outer product of the eigenvalues of the two matrices.