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.
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).