Relation between Cholesky and SVD When we have a symmetric matrix $A = LL^*$, we can obtain L using Cholesky decomposition of $A$ ($L^*$ is $L$ transposed).
Can anyone tell me how we can get this same $L$ using SVD or Eigen decomposition?
Thank you.

If you have the SVD of a positive semi-definite matrix you can easily rewrite this to $L L^*$.  However, this isn't the $L$ the cholesky composition would have computed.
$\begin{align}
A &= U\Sigma V^* && \textrm{SVD def.}\\
U &= V &&\textrm{since A is symmetric} \\
A &= \left(U \sqrt{\Sigma}\right) \left(U \sqrt{\Sigma}\right)^*  && \textrm{note that $\sqrt{\Sigma}$ is easily computed as it's diagonal} \\
A &= L L^* && \textrm{with...}\\
L &= U \sqrt{\Sigma}
\end{align}$
Though this isn't the same $L$ as cholesky computes (since it's not triangular), it does satisty $A = L L^*$ which may be enough to be useful to some.
