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I have a $2n \times 2n$ real anti-triangular (skew-triangular?) block matrix of the form

$$ M = \begin{bmatrix} A & B \\ I_n & O_n \end{bmatrix} $$

where $I_n$ is the $n \times n$ identity matrix and $O_n$ is the $n \times n$ zero matrix. Note that the blocks $A$ and $B$ are also $n \times n$. Do the eigenvalues of $M$ have any specific relationships with the submatrices of $A$, $B$ (or their eigenvalues)?

Any theory or discussion would be helpful.

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2 Answers 2

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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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  • $\begingroup$ If we know from the problem domain that A and B are symmetric and positive definite with a certain form, are there any implications? E.g. can the quadratic formula be applied? $\endgroup$ Mar 11, 2015 at 17:22
  • $\begingroup$ @AlexHirzel The quadratic formula does not work for matrices, short reason is square roots. Here is a question (with more links) discussing it. $\endgroup$
    – adam W
    Mar 11, 2015 at 20:19
  • $\begingroup$ Thank you for your reply! I have found square roots for each of my matrices via a diagonalizing Cholesky factorization technique. My $A$ and $B$ are symbolic, SPD, real-valued. I know for my problem I will have three complex conjugate pairs ($A$ is $3x3$) and I want to solve this problem to find constrains for the entries of $A$ and $B$ as well as the solution $X$. Do you think even with these relatively loose conditions the problem is not fruitful? The link you shared seems to deal with general matrices. $\endgroup$ Mar 12, 2015 at 12:27
  • $\begingroup$ as well as the solution, the eigenvalues* (can't edit comments) $\endgroup$ Mar 12, 2015 at 14:36
  • $\begingroup$ @AlexHirzel Actually it isn't the square roots that would be helpful. I was thinking of the matrix form of the quadratic equation. Sounds to me that is the determinant of it that you want. I am not sure of any constraints that would be useful. $\endgroup$
    – adam W
    Mar 13, 2015 at 1:56
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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 eigenvalue $-1$.

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  • $\begingroup$ Do you agree with my edits? $\endgroup$ May 24, 2023 at 7:03

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