I have a control system with a state matrix $S = -B^{-1} A \in \mathbb{R}^{n\times n}$, where:
- $B$ is a strictly positive diagonal matrix
- $A$ is positive definite $M$-matrix
I know that all the eigenvalues of $S$ are real and strictly negative. I would like to find a bound on $\|e^{Sh} \mathbf{v}\|_{\infty}$ in terms of $h$, where $h > 0$ and $\mathbf{v}$ is an $n\times 1$ vector that is known explicitly ($\mathbf{v}\geq 0$).
I am aware of few methods that allow computing $e^{Sh} \mathbf{v}$ explicitly, but they might be expensive and they probably require knowing $h$ explicitly. What I really want is an expression that bounds the infinity norm and that depends on $h$.
A simple bound is the trivial one:
$$\|e^{Sh} \mathbf{v}\|_{\infty} \leq \|e^{Sh} \|_{\infty} \|\mathbf{v}\|_{\infty} \leq e^{h\|S\|_{\infty}} \|\mathbf{v}\|_{\infty}$$ But this is not very good because $h\|S\|_{\infty} \geq 0$ while all exponential terms in $e^{Sh}$ are of negative numbers (since all eigenvalues are negative and $h > 0$).
My question is: is there a tighter bounds for this kind of matrices? Is there a bound on the infinity norm that depends on the spectral abscissa of $S$? (just like the 2-norm bound?) I would like to avoid diagonalizing $S$ because that would be more expensive that computing the product $e^{Sh} \mathbf{v}$ (I think).
Thank you