The variance-covariance matrix $\Sigma$ is related to the correlation matrix $\varrho$ by the equation
$\Sigma = D \varrho D$, where $D = \sqrt{\text{diag}(\Sigma)}$
Now consider the Cholesky decomposition of $\Sigma$ and $\varrho$, where $\Sigma=LL^T$ and $\varrho = CC^T$.
I want to show that the Cholesky decompositions of these are related by $L = DC$. How do I do this?
This is how far I got:
$\Sigma = D \varrho D \implies LL^T = DCC^TD$
$\implies LL^T = DCC^TD^T$, because $D= D^T$ since it is a diagonal (and thus symmetric) matrix.
$\implies LL^T = DC(DC)^T$
It looks like I am very nearly done. Is it possible to "square root" both sides and get $L = DC$?