The sample covariance matrix $\Sigma$ (chosen because it is symmetric and, therefore, diagonalizable) can be decomposed as
$$\Sigma = \text{diag}(\sigma) C \text{diag}(\sigma)$$ where $\text{diag}(\sigma)$ is a matrix with volatilities along the diagonal, and $C$ is the correlation matrix.
Question
If we compute the matrix exponential of the covariance matrix, $$e^\Sigma = \sum_{k=0}^\infty \frac{1}{k!} \Sigma^k$$ does it too have a unique decomposition?
Attempt
Eigenvectors might indicate a possible answer regarding some link, since
$$\Sigma = PDP^{-1}$$ $$e^\Sigma = Pe^DP^{-1}$$
where $D = \text{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix whose entries are the eigenvalues of $\Sigma$, $e^D = \text{diag}(e^{\lambda_1}, ..., e^{\lambda_n})$, and $P$ contains eigenvectors corresponding to the eigenvalues.
This is unlike my case though because I am not interested in eigenvalues, only the actual matrix, or the matrix exponential of any symmetric matrix. Nevertheless, this is the only decomposition I know of so far, so someone could confirm that it is the only one out there
Suppose $\Sigma = PDP^{-1}$. Look at your formula for $e^{\Sigma}$. Note that $$\Sigma^k = \left(PDP^{-1}\right)^k=PDP^{-1}PDP^{-1}\dots PDP^{-1}=PD^kP^{-1}.$$ So we have $$e^{\Sigma}=\sum_{k=0}^\infty\frac{1}{k!}\Sigma^k=\sum_{k=0}^\infty\frac{1}{k!}PD^kP^{-1}=P\left(\sum_{k=0}^\infty\frac{1}{k!}D^k\right)P^{-1}=Pe^{D}P^{-1}.$$ And because $D$ is diagonal, $D^k$ and hence $e^D$ are really easy to compute: just take $D$, and raise the diagonal entries to the power of $k$ (for $D^k$) or exponentiate them (for $e^D$).