I was reading a research paper and came across the following two claims:
Since any unitary matrix can be represented as the exponential of a skew hermitian matrix, $U$ can be expressed as $e^\Sigma$, where $\Sigma$ is skew hermitian.
The eigenvalues, $\lambda$, of positive definite hermitian $R$ matrix are all greater than zero and can be written as $\lambda=e^{\alpha}$ for some $\alpha\in\mathbb{R}$. Furthermore, $R$ can be written as $R=e^G$, where $\lambda_i(G)=\alpha_i$.
In both cases, author claims that any unitary and positive definite matrix can be written in exponential form. I think that it is not always a case, unless both matrix $U$ and $R$ are diagonal ($\Sigma$ and $G$ as well). There were no assumptions in the paper whether these matrices are diagonal or not, but is it safe to assume that they are?
Let $M$ be a normal matrix. By the spectral theorem it is diagonalizable so
$$ M = UDU^{-1}\\ $$
Suppose $D = e^N$
$$ M = U e^N U^{-1}\\ = U (\sum_{k=0}^\infty \frac{N^k}{k!}) U^{-1}\\ = (\sum_{k=0}^\infty U \frac{N^k}{k!} U^{-1})\\ = (\sum_{k=0}^\infty \frac{(UNU^{-1})^k}{k!})\\ = e^{U N U^{-1}}\\ $$