1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user51196 Jul 28 '16 at 16:18
  • $\begingroup$ Hi. No, I didn't find anything. I asked my teacher and he didn't know either. @noether $\endgroup$ – mehrdad Aug 20 '16 at 17:04
  • $\begingroup$ My problem was answered here, math.stackexchange.com/questions/1874032/… , maybe you can find a hint there $\endgroup$ – user51196 Aug 21 '16 at 14:39
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.