enlarge the separation between two matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:45:44Z http://mathoverflow.net/feeds/question/60723 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60723/enlarge-the-separation-between-two-matrices enlarge the separation between two matrices Federico Poloni 2011-04-05T18:21:59Z 2011-04-06T08:11:16Z <p>The <em>separation</em> 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).</p> <p>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.</p> <p>Let $S$ be a stable matrix, i.e., $\Re\lambda&lt;0$ for each of its eigenvalues $\lambda$. Let $U$ be an unstable matrix, i.e., $-U$ is stable. I would like to prove that $$\operatorname{sep}(S,U)&lt;\operatorname{sep}(S,kU)$$ 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, $$\operatorname{sep}(S,O)&lt;\operatorname{sep}(S,U),$$ 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.</p> <p>Is there any known result in this direction, to your knowledge?</p> http://mathoverflow.net/questions/60723/enlarge-the-separation-between-two-matrices/60786#60786 Answer by S. Sra for enlarge the separation between two matrices S. Sra 2011-04-06T08:11:16Z 2011-04-06T08:11:16Z <p>For the Frobenius norm the answer seems to be <em>no</em>. I haven't tested the operator-2 norm, but it might meet a similar fate.</p> <p>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 $A$.</p> <p>Now, we wish to check whether $$\Delta(S,0) &lt; \Delta(S,U),$$ for a stable $S$ and unstable $U$. In our notation, this inequality amounts to checking if $$\sigma_\min(I\otimes S) = \sigma_\min(S) &lt; \sigma_\min(I\otimes S - U^T\otimes I).$$ Seems like there should be an easy counterexample to this assertion. Below is a brute force numerical example:</p> <p>For simplicity, I try out with $2 \times 2$ matrices. Consider the following matrices: $$S=\begin{bmatrix}-0.4543 &amp; 0.0817\\ -0.6674 &amp; -0.7632 \end{bmatrix},\qquad\qquad U= \begin{bmatrix} 1.1757 &amp; -0.5510\\ 2.2971 &amp;-0.8426 \end{bmatrix}$$</p> <p>We have $Re(\lambda(S)) = (-.6088,-.6088)$, while $Re(\lambda(U)) = (.1666,.1666)$</p> <p>$\sigma_\min(S) = .3837$, while $\sigma_\min(I\otimes S - U^T\otimes I) = .1713$</p> <p>It seems that more meaningful bounds might be possible, if we restrict $S$ and $U$ to be normal matrices.</p>