8
$\begingroup$

The matrix $A$ below is a block diagonal matrix where each block $A_i$ is a $4 \times 4$ matrix with known eigenvalues.

$$A= \begin{pmatrix}A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_n \end{pmatrix}$$

How do I find the eigenvalues of the block diagonal matrix $A$? Does this mean that I will have $4 n$ eigenvalues?

Am I correct in thinking that the eigenvalues of the block diagonal matrix $A$ above are just a list of the individual eigenvalues of each $A_i$ and not the product of everything?

$\endgroup$
  • $\begingroup$ You are right, its the list of eigenvalues of each $A_i$. $\endgroup$ – ForestGump Jun 1 '15 at 15:50
8
$\begingroup$

Since,

$\det(A-\lambda I)=\det(A_1-\lambda I)\det(A_2-\lambda I)...\det(A_n-\lambda I)$,

the eigenvalues of $A$ are just the list of eigenvalues of each $A_i$.

$\endgroup$
1
$\begingroup$

You have a $4n \times 4n$ matrix, so you expect $4n$ eigenvalues (with multiplicities accounted separately).

The eigenvalues will be indeed the eigenvalues of the original submatrices. It's easy to see, say $A_2 \vec{x} = \lambda {x}$. Then look at the block vector $\vec{y} = (0, \vec{x}, 0, \ldots 0)^T$ and note that $A \vec{y} = \lambda \vec{y}$.

$\endgroup$

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.