# Eigenvalues of anti-triangular block matrix

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.

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}

• 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? Mar 11, 2015 at 17:22
• @AlexHirzel The quadratic formula does not work for matrices, short reason is square roots. Here is a question (with more links) discussing it. Mar 11, 2015 at 20:19
• 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. Mar 12, 2015 at 12:27
• as well as the solution, the eigenvalues* (can't edit comments) Mar 12, 2015 at 14:36
• @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. Mar 13, 2015 at 1:56

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 $$-1$$.

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