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I have a 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 matrices each of dimension $ (n \times n)$.

Question is, do eigenvalues of $M$ have any specific relationships with sub-matrices of $A$, $B$ ( or their eigenvalues) ?

Any theory or discussion 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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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? – Alex Hirzel Mar 11 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. – adam W Mar 11 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. – Alex Hirzel Mar 12 at 12:27
as well as the solution, the eigenvalues* (can't edit comments) – Alex Hirzel Mar 12 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. – adam W Mar 13 at 1:56

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