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I have an real anti-triangular matrix

$M=\left[ \begin{array}{cc} A & B \\ I & 0 \\ \end{array} \right]$ where I is an identity matrix. $A$, $B$, $I$, $0$ are all square real matrix with the same dimention $n\times n$.

Qestions is, do eigenvalues of $M$ have specific relationships with submatrix $A$, $B$?

Any theroy or discussions would be helpful.

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2 Answers 2

Nope as your matrix doesn't need to have a single eigenvalue in general, look at the trivial case where $n=1$ and your blockmatrix is \[\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix} \] As the characteristic polynomial is $x^2+1$ it doesn't have any eigenvalues over $\mathbb{R}$ while $A$ hast the eigenvalue $0$ and $B$ has the eigenavlue $-1$.

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The exact relationship boils down to a matrix equation, so it does not simplify further with respect to the individual eigenvalues:

\begin{align} &\operatorname{det}\left(\matrix{xI - A & -B \\ -I & xI}\right) \\ =& \pm \operatorname{det}\left(\matrix{ -I & xI \\ xI - A & -B}\right) \quad\text{sign depending on number of row swaps}\\ =& \pm \operatorname{det}\left(\matrix{ -I & xI \\ 0 & -B + xI(xI - A)}\right) \\ =& \pm \operatorname{det}\left(B - xI(xI - A)\right) \\ =& \pm \operatorname{det}\left(B - x^2I + xI\cdot A\right) \\ \end{align}

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