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

I begin with terminology I use in the question. A real square matrix $A$ is

  • negative-stable if for every eigenvalue $\lambda$ of $A$, ${\mathrm{Re}}(\lambda) < 0$;
  • $\ast$-negative-stable if for every eigenvalue $\lambda$ of $A$, either $\lambda = 0$ or ${\mathrm{Re}}(\lambda) < 0$;
  • nonpositive-stable if for every eigenvalue $\lambda$ of $A$, ${\mathrm{Re}}(\lambda) \leqslant 0$.

I made up the term '$\ast$-negative-stable' and I would welcome better and/or established terminology. For example, the Laplacian matrix of a nonnegatively weighted (directed or undirected) graph is $\ast$-negative-stable.

To put it broadly, I am looking for what is known about matrix operations that preserve the above stability properties.

Let $A$ be a real $n{\times}n$ matrix and let $u$, $v$, $w$ be real $n{\times}1$ vectors. Consider the real $n{\times}n$ matrices $D = \mathrm{diag}(u)$ and $B = vw^{\mathrm{T}}$, and the real number $\alpha = w^{\mathrm{T}}v$. I am particularly interested in what additional conditions on the matrix $A$ would make the following implications true. (I do not mean simultaneously true.) They concern preserving stability from $A$ to $AD$ for the first three and from $A$ to $A+B$ for the last three.

  1. ( $A$ is negative-stable and $u$ is positive ) $\Rightarrow$ ( $AD$ is negative-stable )
  2. ( $A$ is $\ast$-negative-stable and $u$ is nonnegative ) $\Rightarrow$ ( $AD$ is $\ast$-negative-stable )
  3. ( $A$ is nonpositive-stable and $u$ is nonnegative ) $\Rightarrow$ ( $AD$ is nonpositive-stable )
  4. ( $A$ is negative-stable and $\alpha < 0$ ) $\Rightarrow$ ( $A + B$ is negative-stable )
  5. ( $A$ is $\ast$-negative-stable and $\alpha \leqslant 0$ ) $\Rightarrow$ ( $A + B$ is $\ast$-negative-stable )
  6. ( $A$ is nonpositive-stable and $\alpha \leqslant 0$ ) $\Rightarrow$ ( $A + B$ is nonpositive-stable )

In implications 2 and 5 about $\ast$-negative-stability, it would be acceptable to assume that $A$ is similar to a Laplacian matrix (but Laplacian matrices should not be assumed to be symmetric). Would that be sufficient?

Addendum 1

Here is a way $\ast$-negative stability can be useful in studying negative/Hurwitz stability. Suppose I know that a matrix $A$ is similar to $C = \begin{pmatrix} O_{p{\times}p} & O_{p{\times}q} \\\ S & T \end{pmatrix}$, where $T$ is nonsingular. Then $T$ is negative-stable if and only if $A$ is $\ast$-negative-stable, a potentially useful observation if $A$ looks easier to work with than $T$. The notion of nonpositive stability can become useful in similar (but not identical) circumstances.

share|improve this question
1  
I suggest you to look at M-matrix theory, if you haven't yet. It should help you settling 1--3, at least. Moreover, I'd like to point out that our second and third definition of stability are somehow counterintuitive: for instance, $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ is "*-negative-stable" and "nonpositive-stable", yet $e^{At}$ diverges. Usually a condition on the multiplicities of purely imaginary eigenvalue is assumed in addition. –  Federico Poloni May 16 '12 at 7:39
    
Thanks, Federico. I recognize that "$\ast$-negative-stable" and "nonpositive-stable" can seem strange without explanation of where they come from. Also, note that $\ast$-negative-stable matrices do not have purely imaginary eigenvalues, except possibly zero. I augmented my question with Addendum 1 to provide some motivation. –  Gilles Gnacadja May 16 '12 at 17:32
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.