Suppose we have a covariance matrix $\Sigma$. I know that the Cholesky decomposition $A^T A$ can be found from the LDL decomposition using $$ \Sigma = LDL^T = (LD^{\frac 1 2})(LD^{ \frac 1 2 })^T = A^TA $$.

I was wondering if you can go the other way -- find the LDL decomposition from the Cholesky, I couldn't see how to do it!

up vote 5 down vote accepted

Let $S=D^{\frac12}$, so that $S$ is a diagonal matrix with diagonal $s$.

Note that when you compute $A=LS$, multiplying $L$ by $S$ multiplies each column of $L$ by the corresponding element of $s$, and that $L_{ii}=1$. So you need to find the $s_i$ that would give $A_{ii}/s_i=1$, and then divide every element of the $i$th column of $A^\top$ (which is the $i$th row of $A$) by $s_i$ to give the $i$th column of $L$.

Then the diagonal of $D$ is trivial to compute from $S$.

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