enlarge the separation between two matrices - MathOverflow

enlarge the separation between two matrices

2011-04-05T18:21:59Z

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?

Answer by S. Sra for enlarge the separation between two matrices

2011-04-06T08:11:16Z

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.