Logarithm of the determinant of a positive definite matrix
By Mia Morrison •
For positive definite $C=LL^T$, where $L$ is the lower triangular Cholesky factor of $C$, why is $\log(\det(C))=2\operatorname{trace}(\log(L))$? I know that if $\{\lambda_i\}$ are the eigenvalues of $C$, $\det(C)=\prod_i\lambda_i$, so that $\log(\det(C))=\sum\log(\lambda_i)$ but I'm not sure where to go from there.
$\endgroup$ 31 Answer
$\begingroup$Hint 1: $\det(C)=\det(LL^T)=\det(L)\det(L^T)=\det(L)^2$, so $\log\det(C)=2\log\det(L)$. Denote by $\lambda_i$ the eigenvalues of $L$ and continue in the same way as you tried.
Hint 2: For the Jordan normal form $L=SJS^{-1}$ it holds $\log(L)=S\log(J)S^{-1}$, so $$ \operatorname{trace}\log(L)=\operatorname{trace}(S\log(J)S^{-1})=\operatorname{trace}\log(J). $$
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