Exponential of irreducible, aperiodic Hurwitz/Metzler matrix is irreducible or has no zero entries?

Question

I am working with a Hurwitz-stable, Metzler matrix $A$ with nonpositive diagonal ($A_{ii}\leq0$ for all $i$) and nonegative off diagonal ($A_{ij}\geq0$ for all $i\neq j$). I want its exponential to be either strictly positive ($e^A>0$) or at least nonnegative and irreducible ($e^A\geq0$ and irreducible).

I have a feeling that $e^A>0$ in this situation actually. I know that it is nonnegative by a basic argument using Lie product formula with separating out part of the diagonal and leading the rest as an irreducible Markov generator:

$$A= D + G$$

Where $D$ is a diagonal matrix with potentially a mix of positive, zero, and negative entries on its diagonal, and $G$ is a Markov generator matrix. Then

$$ e^A = \lim_{n\to\infty}(e^{\frac1n D} e^{\frac1n G})^n$$

and the matrix product on the right is strictly positive for each $n$. This shows $e^A$ is nonnegative.

Of course the matrix on the right is also irreducible for each $n$, but that doesn't necessarily tell me $e^A$ is irreducible. Right?

After the comments below, I thought maybe something about a one parameter matrix group might help, but I couldn't figure anything out.

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Edit: I made the question more focussed.

Edit 2: Originally, I was interested in when $e^A$ is irreducible for an irreducible $A$ (you can see the original wording in the edit history). Through a discussion in the comments, I was able to basically answer both of my original questions. See the answers posted below for two different approaches.

Answer 1

Ok, I have an answer thanks to the discussion in the comments with @user8675309.

Let $A$ be a real square matrix that is nonzero off the diagonal and is irreducible. The diagonal entries can be any real numbers. Then $e^A$ is strictly positive.

We can choose a positive number $\delta$ large enough to make $A+\delta I$ (with $I$ the identity matrix) a nonnegative matrix. It is also irreducible since $A$ is, and it is aperiodic due to having a nonzero entry in the diagonal (that plus irreducibility gives aperiodicity). Hence we know that $e^{A+\delta I}$ has only nonnegative entries. In fact $A+\delta I$ is primitive since it is nonnegative, irreducible, and aperiodice, and this shows that $(A+\delta I)^n$ will be strictly positive for all $n$ large enough. This tells us that $e^{A+\delta I}$ is in fact strictly positive.

Now we show that $e^A$ is strictly positive. Since $A$ and $I$ commute, we have $e^{A+\delta I}=e^A e^{\delta I}$. Now we have, for the $ij$ entry of $e^{A+\delta I}$, that $(e^{A+\delta I})_{ij}=(e^A)_{i\cdot}(e^{\delta I})_{\cdot j}$ where we mean the $i^{th}$ row and $j^{th}$ column on the right side. But we can show that this simplifies to:

$$\underbrace{(e^{A+\delta I})_{ij}}_{>0}=\underbrace{(e^A)_{ij}}_{?} \cdot \underbrace{e^\delta}_{>0}.$$ Finally, this shows that $e^A$ can only have positive entries everywhere.

Also, we get some sense on what the answer to my more general question is. Let matrix $A$ have only nonnegative entries off the diagonal (but not necessarily irreducible nor aperiodic). By the above argument, we can assume its diagonal is strictly positive since the diagonal entries of $A$ cannot contribute to the sign of the entries for $e^A$ (nor whether they are zero). In other words, $e^{A+\delta I}$ is nonnegative for large enough $\delta$ implies that $e^A$ is nonnegative for $A$ with arbitrary real diagonal and nonnegative off diagonal. This automatically shows that the question of periodicity is not important as well---it's not hard to come up with a periodic example with $e^A$ strictly positive. Letting negative numbers off the diagonal can completely break this. See: Exponential of the zero matrix

We have that $e^A$ is nonnegative and $A^n$ is nonnegative for all $n$. Where can it have a zero? This would require: $$0=(e^A)_{ij}=I_{ij}+A_{ij}+\frac12 (A^2)_{ij}+\cdots$$ which means $i\neq j$ is required and $(A^n)_{ij}=0$ for all $n$. This means that the matrix must be reducible. There must be an index $i$ and index $j$ such that there is no path in the adjacency graph from $i$ to $j$.

So we summarize:

If $A$ is nonnegative off the diagonal, then:

• $e^A$ is strictly positive if and only if $A$ is irreducible, and
• $e^A$ will have some (off-diagonal) zero entries if and only if $A$ is reducible.

This could probably be made rigorous using a permutation similarity transform. A reducible matrix is permutation-similar to an upper triangular block matrix (i.e. by just swapping rows and columns). The exponential of such an upper triangular block matrix will always have zeros in the lower off diagonal, and a permutation transform just moves those zeros around without eliminating them.

Answer 2

For the record here is a proof that I originally hinted at. $A=G+D$ with $D$ a real diagonal matrix and $G$ non-negative, zero on the diagonal and irreducible.

1.) The Lie Product Formula shows that $\exp\big(A\big)$ is a non-negative matrix. So we may use Perron-Frobenius Theory and confirm it is irreducible and aperioidic by examining its sprectrum.

2.) The diagonal (self loop) information contained by $D$ is in some sense a distraction. The fact that $G$ is real non-negative and irreducible gives us a lot of information. In particular it tells us that $A$ has a single positive (Perron) eigenvector $\mathbf v_1$ and the associated eigenvalue $\lambda_1$ is simple. Proof: apply Perron-Frobenius theory to $B:=A+\delta I$ for large enough $\delta \gt 0$.

3.) We know $B$ has no other eigenvalue with greater modulus than the Perron root and that the Perron root is positive. This means the Perron root of $B$ has a strictly maximal real component $\implies A$'s Perron root has a strictly maximal real component, i.e. $Re(\lambda_1)\gt \lambda_j$ for $j\neq 1$.

Combining this with the fact that $\big \vert \exp(z)\big \vert = \exp\big(Re(z)\big)$ tells us that $\vert\exp(\lambda_1)\vert\gt \vert\exp(\lambda_j)\vert$ for $j\neq 1$. And of course $\exp\big(A\big)\mathbf v_1 = e^\lambda_1 \mathbf v_1$. Conclude that real non-negative $\exp\big(A\big)$ has positive Perron vector $\mathbf v_1$ with simple Perron root $e^{\lambda_1}$ and all other eigenvalues have strictly smaller modulus. The former proves $\exp\big(A\big)$ is irreducible and the latter proves aperiodicity.