Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The separation between two square matrices $A$ and $B$, often used as a measure of the sensitivity of invariant subspace problems, is defined as $$ \operatorname{sep}(A,B)=\min_{X\neq 0}\frac{\left\Vert AX-XB\right\Vert}{\left\Vert X\right\Vert} $$ for a suitable matrix norm (see e.g. the book by Stewart and Sun or Golub and Van Loan, Section 7.2.4).

Intuitively, it measures the distance between the spectra of $A$ and $B$: if they share an eigenvalue, $\operatorname{sep}(A,B)=0$, and if at least two of their eigenvalues are very close then it is small.

Let $S$ be a stable matrix, i.e., $\Re\lambda<0 $ for each of its eigenvalues $\lambda$. Let $U$ be an unstable matrix, i.e., $-U$ is stable. I would like to prove that $$\operatorname{sep}(S,U)<\operatorname{sep}(S,kU)$$ for $k>1$, or at least some weaker result on the lines of "if I take the eigenvalues more far apart than they are, then the separation increases". For instance, $$\operatorname{sep}(S,O)<\operatorname{sep}(S,U),$$ where $O$ is the zero matrix (of size $1\times 1$, or of the same size of $U$, does not matter) would suit my needs. Establishing this result for at least one among Euclidean and Frobenius norm would be fine.

Is there any known result in this direction, to your knowledge?

share|improve this question
4  
I don't know anything about separation of matrices, but the eigenvalues of $kU$ can be much closer to the eigenvalues of $S$ than the eigenvalues of $U$ with your assumptions. Eg, for some small $\epsilon$, you could have $S$ with an eigenvalue at $i - \epsilon$, $U$ with an eigenvalue at $(i + \epsilon)/2$, and $k=2$. This doesn't seem to fit your intuition. Am I misunderstanding what a "stable matrix" is? –  Dylan Thurston Apr 6 '11 at 2:16
add comment

1 Answer 1

up vote 3 down vote accepted

For the Frobenius norm the answer seems to be no. I haven't tested the operator-2 norm, but it might meet a similar fate.

Here's a simplistic argument. For the Frobenius norm, the separation as defined above reduces to $$\Delta(S,U) := \text{sep}(S,U) = \sigma_\min( I \otimes S - U^T\otimes I),$$ where $\sigma_\min(A)$ denotes the minimum singular value of the matrix $A$.

Now, we wish to check whether $$\Delta(S,0) < \Delta(S,U),$$ for a stable $S$ and unstable $U$. In our notation, this inequality amounts to checking if $$\sigma_\min(I\otimes S) = \sigma_\min(S) < \sigma_\min(I\otimes S - U^T\otimes I).$$ Seems like there should be an easy counterexample to this assertion. Below is a brute force numerical example:

For simplicity, I try out with $2 \times 2$ matrices. Consider the following matrices: $$S=\begin{bmatrix}-0.4543 & 0.0817\\\\ -0.6674 & -0.7632 \end{bmatrix},\qquad\qquad U= \begin{bmatrix} 1.1757 & -0.5510\\\\ 2.2971 &-0.8426 \end{bmatrix}$$

We have $Re(\lambda(S)) = (-.6088,-.6088)$, while $Re(\lambda(U)) = (.1666,.1666)$

$\sigma_\min(S) = .3837$, while $\sigma_\min(I\otimes S - U^T\otimes I) = .1713$

It seems that more meaningful bounds might be possible, if we restrict $S$ and $U$ to be normal matrices.

share|improve this answer
    
Thanks for the counterexample! Not the answer I hoped for, but still a good answer, and points me to the right direction --- probably I need to exploit some additional properties of the involved matrices if I want to get an useful result. –  Federico Poloni Apr 7 '11 at 9:30
    
You're welcome! But the goal of developing a monotonic measure of separation is of course a very important. Good luck. –  Suvrit Apr 8 '11 at 7:35
add comment

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.