Let $C$ be a $2 \times 2$ matrix with real entries, and $x\in\mathbb{R}^2$. We write $x > 0$ if both coordinates are strictly positive.

Suppose $x>0$, under what conditions on $C$ and $x$ do we also have $e^Cx>0$? Note that the inequalities are strict.

(In my example, $C$ is negative-definite, but don't use that if you don't need to.)

  • $\begingroup$ $e^C$ is a two-by-two matrix. How are you comparing it to a two-by-one vector? $\endgroup$ – Calculon Sep 11 '14 at 0:00
  • $\begingroup$ Thanks, saw the typo just as you did. Fixed. $\endgroup$ – jlperla Sep 11 '14 at 0:02
  • $\begingroup$ Is this equivalent to asking when is the matrix exponential of a negative definite matrix negative definite, or not quite? $\endgroup$ – jlperla Sep 11 '14 at 0:03
  • $\begingroup$ no it is not equivalent to that. if you express $e^C$ in terms of eigenvalues and eigenvectors of $C$, that might lead you to the answer. $\endgroup$ – Calculon Sep 11 '14 at 0:05
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    $\begingroup$ Your question is confusingly formulated. The title asks whether matrix exponentials can ever be negative, and at the end you mention the matrix is negative definite. The answer (with the given explanation of what you mean by "negative") is the same either way, namely "yes they can be negative", but this makes it hard to properly answer your question, since one is tempted to spend a lot of time on the additional restrictive condition, which is in fact irrelevant to the main question. $\endgroup$ – Marc van Leeuwen Sep 11 '14 at 4:54

The strict inequality is trickier but for $e^C x \geq0 $ to hold with any $x\geq0$ the matrix $e^C$ must have only non-negative entries. This is different from positive or negative definiteness, there is no relation in either direction. For $C$ it is sufficient to have non-negative entries off-diagonal. This is because $e^{tC}=I+tC+o(t)$, so for small positive $t$ the identity $I$ dominates on the diagonal, and the $C$ entries dominate off the diagonal. For $t=1$ we use the exponential property $e^{C}=(e^{\frac1nC})^n$ to see that positivity still holds. Indeed, $e^{\frac1nC}$ has non-negative entries for large $n$, and this is preserved under matrix multiplication.

The non-negative off-diagonal is also necessary if we ask to have $e^{tC}x\geq0$ for all numbers $t\geq0$ (and all $x \geq0 $). If $C$ has negative off-diagonal entries then for small $t$ we have $e^{tC}\approx I+tC$ and off-diagonal entries of $A:=e^{tC}$ will also be negative. Let's say $a_{12}<0$. Multiplying it by $x=(\varepsilon, R)^T$ with small enough $\varepsilon>0$ and large enough $R>0$ we can ensure that the first coordinate of $Ax$ is negative.

To have $e^C x>0$ for all $x>0$ it is sufficient, by the same reasoning as above, that $e^C$ have strictly positive entries, and that $C$ correspondingly have strictly positive entries off-diagonal. But it can probably be somewhat relaxed.

  • $\begingroup$ Thank you. This answer makes me think that it makes sense to ask the full question, and that I stripped things down too far. See math.stackexchange.com/questions/927141/… $\endgroup$ – jlperla Sep 11 '14 at 4:15
  • $\begingroup$ Thanks for the answer which I accepted. I have posted a variation which I realize is much cleaner and what i am really getting at. Take a look at: math.stackexchange.com/questions/929113/… and note that the negative case tends to happen for very large $t$. $\endgroup$ – jlperla Sep 12 '14 at 21:00

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