Matrix exponential: $e^A\ge 0\iff a_{ij}\ge 0$ when $i\neq j$

Question

Let $A$ be a $n$ by $n$ matrix. Prove that that $$a_{ij}\ge 0 \text{ whenever }i\neq j\iff e^A\text{ has all entries }\ge 0.$$

I'd like just a hint for now please.

Sufficiency part easily follows from the series expansion of $e^A$. But I think necessity part is not trivial, also may not be true. — obareey, Dec 13 '13 at 18:46

user1551 · Accepted Answer · 2013-12-13T19:26:22.070

Since $e^A=e^{-kI}e^{A+kI}$, the forward implication is trivial if you consider the series expansion of $e^{A+kI}$ for sufficiently large $k>0$ (so that $A+kI$ is entrywise nonnegative).

The backward implication is false. Let $J$ be the $3\times3$ Jordan block corresponding to the eigenvalue zero and let $A=I+J-\varepsilon J^2$ for some small $\varepsilon>0$. Then the $(1,3)$-th entry of $A$ is $-\varepsilon<0$, but $$ e^A=e^Ie^Je^{-\varepsilon J^2}=e\left(I+J+\frac12J^2\right)(I-\varepsilon J^2) =e\left[I+J+\left(\frac12 - \varepsilon\right) J^2\right] $$ is entrywise nonnegative.

score 7 · Answer 2 · answered Dec 13 '13 at 19:42

7

There's already a counter-example by user1551, but here's an even simpler one.

If we let $$A=\left[\matrix{0&1\\-1&0}\right],$$ then $e^{\theta A}$ rotates the plane by an angle $\theta$. Enter $\theta=2\pi$ and you get $e^{2\pi A}=I$ where $I$ is the identity matrix which has non-negative elements.

answered Dec 13 '13 at 19:42

Einar Rødland

9,707

1

(You are rotating clockwise, though.) – Andrés E. Caicedo Dec 13 '13 at 19:55
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@Andres That depends on whether the matrix is acting on the right (i.e. on row vectors) or on the left (on column vectors). The former is counterclockwise, while the latter is clockwise. – Einar Rødland Dec 13 '13 at 20:09
Yes. Nice example, too. – Andrés E. Caicedo Dec 13 '13 at 20:14

Matrix exponential: $e^A\ge 0\iff a_{ij}\ge 0$ when $i\neq j$

2 Answers2