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This question was somewhat answered here: Fast trace of inverse of a square matrix. However, I feel like there was no complete answer wrt the Cholesky case.

I have the matrix $\Sigma=LL^T$. Is there a way of getting $Tr(\Sigma^{-1})$ without using the SVD? I'm guessing eigen decomposition is just as costly as SVD. I have already computer the lower Cholesky matrix $L$ for a previous computation.

The matrix is symmetric, positive definite and (unfortunately) dense.

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What about computing an upper triangular Cholesky factor instead? Then, from the shape of $\Sigma^{-1}=U^{-T}U^{-1}$, you get that the diagonal entries are the squares of the inverse of the diagonal of $U$. – Federico Poloni Apr 15 '14 at 6:20
Oh, and for a SPD matrix, essentially the eigendecomposition is the SVD. – Federico Poloni Apr 15 '14 at 6:21
Theoretically, yes. Unfortunately my matrix has some eigen values close to zero that sometimes it dips below zero. Basically numerical problems. For the first comment, I guess I was hoping not to use inverse of U (can do the same operation with L) – Sachin_ruk Apr 15 '14 at 7:47
So you wish to compute $\mbox{trace}(L^{-T}L^{-1})$ without actually computing $L^{-1}$? You can use quadrature + randomisation based algorithms to approximate this term, but I don't know if that's what you're after.... – Suvrit Apr 15 '14 at 10:05

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