Is it true - $\lim_{t\rightarrow \infty}e^{At}=\mathbf{0}_{n\times n}?$

Question

Let $A_{n\times n}$ be a diagonally dominant matrix such that each diagonal entry is negative and of absolute value strictly less than $1$, while each nondiagonal entry is nonnegative. Then is it true that $$\lim_{t\rightarrow \infty}e^{At}=\mathbf{0}_{n\times n}.$$ I expect it is true, but I do not have a proof.

Is it true that $A$ is negative definite?

Sorry, I have left one more condition that modulus of all the entries is less than 1. — G_0_pi_i_e, Oct 12 '15 at 14:41
"Strictly less than one" is not important (can be done by scaling). In short: simply diagonally dominant (non-strict inequality) is not enough. For strictly diagonally dominant matrices it is true. If $A$ is Hermitian then it is also true that it is negative definite. — A.Γ., Oct 12 '15 at 20:05
@A.G.! Could you please explain - how is it true for a strictly diagonally dominant matrix? — G_0_pi_i_e, Oct 13 '15 at 02:54

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

For simply diagonally dominant it is not true. The counterexample is $$ A=\left[\matrix{-1 & 1\\1 & -1}\right]. $$

For strictly diagonally dominant: assume that the matrix $A=\{a_{ij}\}$ satisfies

$a_{ij}\ge 0$, $i\ne j$
$a_{ii}+\sum_{j\ne i}a_{ij}<0$

then by Gerschgorin's theorem all the eigenvalues of $A$ have negative real parts, i.e. the matrix $A$ is stable, therefore, the result follows (see here or here).

Is it true - $\lim_{t\rightarrow \infty}e^{At}=\mathbf{0}_{n\times n}?$

1 Answers1