# 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?

• Hi, I am also looking for a similar answer. Did you have any progress on that formula? or at least bounds for the eigenvalues or something? – user51196 Jul 28 '16 at 16:18
• Hi. No, I didn't find anything. I asked my teacher and he didn't know either. @noether – mehrdad Aug 20 '16 at 17:04
• My problem was answered here, math.stackexchange.com/questions/1874032/… , maybe you can find a hint there – user51196 Aug 21 '16 at 14:39

## 1 Answer

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?