# Can someone look at my proof about the convergence of $e^{-tA}$

Hi I am trying to prove that if A is a symmetric positive definite matrix then $e^{-tA}\rightarrow 0$ as $t\rightarrow\infty$.

So I have attempted an answer but I'm not sure it is correct.

$e^{-tA}=\sum_1^{\infty}(-1)^n\frac{t^nA^n}{n!}$ so set $a_n(t)=\frac{t^nA^n}{n!}$ then by the operator norm

$a_n(t)\leq \frac{(||A||t)^n}{n!}\rightarrow 0$ for all t so by the alternating series test $e^{-tA}\rightarrow 0$ as t tends to infinity.

I haven't use any properties of the matrix A which is one of the reasons I think it is a bit dodgy although one of the following questions is;

Is the result still true for general positive definite A? Rigorously justify your answer.

So I'm pretty confused.

• You have proved (assuming you just forgot to write the norms properly) that the matrix exponential is well defined for every fixed $t$. As @ivanpenev answered, you can prove that $||e^{-tA}||\leq Ke^{-t\mu}$ for some $K>1$ and every $\mu$ greater than the sum of all eigenvalues of $A$.
– Koto
Commented Jul 25, 2017 at 16:34

The alternating series test applies to series of real numbers, not to matrices. In order to prove the statement, one could argue as follows: Since $A$ is symmetric, there exists an orthogonal matrix $T$ such that $D = T^{-1}AT$ is a diagonal matrix with real entries (the eigenvalues of $A$). Since $A$ is positive definite, all eigenvalues $\lambda_1$, $\ldots$, $\lambda_n$ of $A$, and hence of $D$, are strictly positive. Therefore, $$e^{-tA} = Te^{-tD}T^{-1} = T\begin{bmatrix}e^{-\lambda_1 t} & \phantom{0} & \phantom{0} \\ \phantom{0} & \ddots & \phantom{0} \\ \phantom{0} & \phantom{0} & e^{-\lambda_n t}\end{bmatrix} T^{-1} \to 0 \quad \text{as} \quad t \to \infty.$$ I'm not an expert on the subject, but I'm not aware of the notion of positive definiteness of matrices being defined for non-symmetric or non-hermitian matrices (in the complex case).