# Decomposition of the matrix exponential (of the covariance matrix)

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

$$\Sigma = PDP^{-1}$$ $$e^\Sigma = Pe^DP^{-1}$$
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.
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 $$e^D$$).