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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?

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  • $\begingroup$ You are right, its the list of eigenvalues of each $A_i$. $\endgroup$
    – ForestGump
    Commented Jun 1, 2015 at 15:50

2 Answers 2

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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 $A_i$.

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  • $\begingroup$ Which book can I cite when mentioning this, or is it common knowledge? $\endgroup$
    – Isi
    Commented May 15, 2023 at 12:58
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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 $A \vec{y} = \lambda \vec{y}$.

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