min value on the diagonal of matrix exponential

Question

I'm trying to compute the minimum value along the diagonal of $\exp(-A)$ efficiently: $$\min \mathrm{diag}(\exp(-A))$$ Where $A$ is positive semidefinite. Is there a way of relating this quantity to some norm of $A$, or to a function of its eigenvalues? If not, can I bound it?

Answer 1

Let $m=min(diagonal(e^{-A}))$ and $spectrum(A)=\lambda_1\geq\cdots\geq \lambda_n\geq 0$. There is $i$ s.t. $m=e_i^Te^{-A}e_i$. Thus $e^{-\lambda_1}\leq m\leq e^{-\lambda_n}$.

Also, one has $m\leq \dfrac{1}{n}tr(e^{-A})=\dfrac{1}{n}\sum_ie^{-\lambda_i}$.