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Matrix exponential of a positive matrix

Mathematics Asked on December 5, 2021

Let $M$ be an $ntimes n$ real matrix with strictly positive elements, i.e. $M_{ij}>0, forall 1leq i,jleq n$, and let $lambda_1$ be its Perron-Frobenius eigenvalue (i.e. the eigenvalue with largest magnitude, must be real and positive). Is the following inequality always correct?
$$[e^{M}]_{ij}leq e^{lambda_1}, forall 1leq i,j leq n$$

It is easy to prove this when $M$ is symmetric, since when $M$ is symmetric, we have $|M|=lambda_1$, therefore
$$[e^{M}]_{ij}leq |e^{M}|leq e^{|M |} = e^{lambda_1}, forall 1leq i,j leq n.$$
When $M$ is not symmetric, I checked several cases numerically and the answer seems to be true, but I don’t know how to prove this. [The above proof fails because $|M|neq lambda_1$ when $M$ is not symmetric.]

One Answer

This isn't true.

Suppose the claim was true:
In particular, consider positive stochastic matrix $M$ and
$Sigma:= begin{bmatrix}sigma_{1}& mathbf 0^T\ mathbf 0 &mathbf I_{n-1} end{bmatrix} $ with $sigma_1 gt 1$

$e^M$ is a positive matrix and has Perron Root $e$ thus

$sigma_1cdottext{positive constant} = big[Sigma e^M Sigma^{-1}big]_{1,2} = big[e^{Sigma MSigma^{-1}}big]_{1,2} leq e$
but the upper bound is fixed while the lower bound may be made arbitrarily large by selecting large enough $sigma_1$-- which is a contradiction.

working backwards, the counter example is
$M':= Sigma MSigma^{-1}$
with sufficiently large $sigma_1$

Answered by user8675309 on December 5, 2021

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