Does a real symmetric matrix is a positive definite real symmetric matrix by exponential map？

Question

If $A$ is a real symmetric matrix, $\forall r\in \mathbb{R}$, can we have $\exp(rA)$ is a positive definite real symmetric matrix?

Answer 1

Yes. Write $A = P D P^{-1}$ with $D$ diagonal and $P^{-1} = P^T$. Then obviously \begin{equation} \exp(r A) = P \exp(r D) P ^{-1} \end{equation} is symmetric and positive definite because if $D = \operatorname{diag}(\lambda_1,\cdots,\lambda_n)$ then \begin{equation} \exp(r D) = \operatorname{diag}(\exp(r\lambda_1),\cdots,\exp(r\lambda_n)) \end{equation}