Eigenvalues of a tridiagonal block matrix When a tridiagonal matrix is also Toeplitz, there is a simple closed-form solution for its eigenvalues, being $$\lambda_k= a + 2 \sqrt{bc} \, \cos(k \pi / {(n+1)}) , fork=1,...,n$$. Now my question, is there formula for eigenvalues of a tridiagonal block  matrix as well? for example I have $$
A=\left[        
\begin{matrix}
        B & I & 0 \\
        I & B & I \\
        0 & I & B \\
        \end{matrix}
\right]
$$
which
$$
 B=\left[
        \begin{matrix}
        -4 & 1 & 0 \\
         1 &-4 & 1 \\
         0 & 1 & -4 \\
        \end{matrix}
\right]
$$
and $I$, $0$ are 3 by 3 matrices. can the eigenvalues be calculated from a similar formula?

Write $$C:=\left[
        \begin{matrix}
        0 & 1 & 0 \\
         1 &0 & 1 \\
         0 & 1 & 0\\
        \end{matrix}
\right],$$
so that 
$$A=B\otimes I + I\otimes C.$$
Now $B$ and $C$ are symmetric and diagonalisable by orthogonal $\Omega$, $\Delta$ and you tell us you have a formula for the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ of $B$ and $\mu_1, \mu_2, \mu_3$ of $C$. 
Now $\Omega\otimes \Delta$ will simultaneously diagonalise $B\otimes I$ and $I\otimes C$, and hence $A$. Moreover the eigenvalues of $A$ are now patently the nine $\lambda_i+\mu_j$.
Or have I done something very silly?
