Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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 $-1$.

share|improve this answer

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}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.