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Is there any simple relation between the eigenvalues (or the characteristic polynomials) of two matrices $A$ and $B$ with that of matrix $C$ defined as $$ C=\begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix}. $$ $A$ and $B$ are invertible $n\times n$ matrices.

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2 Answers 2

Given two vectors $u,v \in \mathbb{R}^n$ let $w=\begin{bmatrix} u \\ v \end{bmatrix}\in\mathbb{R}^{2n}$ be the vector whose first $n$ components are those of $u$ and whose last $n$ are those of $v$.

The matrix-vector product $C\cdot w$ gives \begin{equation*} \begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} Av \\ Bu \end{bmatrix} \end{equation*} Therefore, solving $Cw = \lambda w$ reduces to solving the following \begin{equation*} \begin{cases} Av = \lambda u \\ Bu = \lambda v \end{cases} \end{equation*} Since $A$ and $B$ are invertible, it follows that $v=\lambda A^{-1}u$ and $u=\lambda B^{-1}v$: \begin{equation*} \begin{cases} Av = \lambda \cdot \lambda B^{-1}v \\ Bu = \lambda \cdot \lambda A^{-1}u \end{cases} \rightarrow \begin{cases} BAv = \lambda^2 v \\ ABu = \lambda^2 u \end{cases} \end{equation*} We conclude that $\lambda\in\mathbb{C}$ is an eigenvalue of $C$ iff $\lambda^2$ is an eigenvalue of $AB$ (or $BA$, since they coincide).

(By the way, this solution covers only the case in which $A$ and $B$ are invertible matrices, but as shown by user1551 they need not be)

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We have $$ \det (C-\lambda I)= \det\begin{bmatrix}-\lambda I&A\\B&-\lambda I\end{bmatrix} = \det(\lambda^2 I -AB). $$ Therefore the eigenvalues of $C$ are the square roots of the eigenvalues of $AB$. So, they depend on the eigenvalues of $AB$ rather than the individual eigenvalues of $A$ and $B$.

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