Matrix exponentials

Question

Let $A(x)$ be a real valued matrix with $x$-dependent coefficients where $x\in \mathbb{R}$. What is the necessary and sufficient condition on $A(x)$ such that the matrix exponential $$\exp( - A(x))$$ is bounded?

I think it should be that there is some constant matrix $B$ so that $A(x) +B$ is a positive matrix or is positive definite. This comes down to what is going on in the non-diagonal terms and I can't see how to deal with it.

Thanks :)

Answer 1

$x\rightarrow e^{-A(x)}$ is bounded iff $g:x\rightarrow tr(e^{-A(x)}e^{-A^T(x)})$ has an upper bound iff the greatest singular value of $e^{-A(x)}$ has an upper bound; you cannot say anything more except if, for example, $A(x)$ is normal; then $g(x)=tr(e^{-(A+A^T)(x)})=\sum_i e^{-\sigma_i(x)}$ where $spectrum(A+A^T)=(\sigma_i)$. Thus $g(x)$ has an upper bound iff there is $M>0$ s.t., for every $i,x$, $-\sigma_i(x)\leq M$.

Answer 2

Suppose that $A(x)$ is bounded, then you have $\|e^{-A(x)}\| \leq e^{\|-A(x)\|}$, therefore is bounded.