# Matrix Operations Preserving Hurwitz Stability

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?

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.
• 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. May 16, 2012 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. May 16, 2012 at 17:32