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For $X=\left(\begin{array}{cc} A & B\\ C & 0\end{array}\right)$, how are eigenvalues of $X$ related to the eigenvalues of $A$?

Rodrigo de Azevedo
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Shyam
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2 Answers2

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Not much can be said. However, if $A$ is square and $X$ is Hermitian (hence $A$ is Hermitian and $C=B^\ast$) and $\lambda_1(M)\le\lambda_2(M)\le\lambda_3(M)\cdots$ denote the eigenvalues of a Hermitian matrix $M$ arranged in increasing order, we have the following interlacing inequality: $$ \lambda_k(X)\le\lambda_k(A)\le\lambda_{k+n-r}(X) $$ for $1\le k\le r$, when $A$ is $r\times r$ and $X$ is $n\times n$.

Also, if the four blocks have equal sizes, the characteristic polynomial of $X$ can be simplified as follows: $$ \det\pmatrix{xI-A&-B\\ -C&xI} = \det(x^2I - xA - BC). $$ Such simplification is valid because the two blocks at the bottom of the LHS commute. You can see that $\det(x^2I - xA - BC)$ has little resemblance to $\det(\lambda I-A)$ and we don't expect any relationship between the eigenvalues of $X$ and $A$ in general.

user1551
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  • What if all the eigenvalues of X are strictly negative but X is not Hermitian, will that tell any thing about A? (If X was Hermitian, eigenvalues of A would have been strictly negative according to the inequality) – Shyam Jun 05 '13 at 04:50
  • @Shyam Almost nothing, except that eigenvalues of $A$ (when $A$ is a square matrix) must sum up to a negative number, because $A$ and $X$ have the same trace. $A$ can have positive or even nonreal eigenvalues. – user1551 Jun 05 '13 at 05:25
  • Thanks a lot! This question is related to a problem of block inverse that I am trying to solve, can you help me with that http://math.stackexchange.com/questions/411492/inverse-of-a-block-matrix – Shyam Jun 05 '13 at 14:45
  • What is $n$? Is that the number of unique eigenvalues? In other words, do you arrange the eigenvalues with multiplicity? – Eric S. Jul 04 '16 at 09:39
  • But then in the interlacing inequality, $n$ is equal to either the size of $A$ or the size of $X$? Come to think of it, what is the size of $X$? By the way, thanks for replying so quickly on a question from 2 years ago. – Eric S. Jul 04 '16 at 13:52
  • @EricS. Please see my new edit. – user1551 Jul 04 '16 at 14:36
  • Thanks for editing! And once again, thank you for your quick responses :) – Eric S. Jul 05 '16 at 06:43
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Say that for some v, $ Av = \lambda v $ and $ C v = 0 $. Then letting $ w = \left(\begin{array}{c} v \\ 0 \end{array}\right) $, what is $ X w $? Similarly if $ w = \left(\begin{array}{c} x \\ y \end{array}\right) $ is an eigenvalue of $ X $ and $ B y = 0 $, then what is $ A x$?

user81021
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