Determinant of $I + LL^T$

2018-03-15 19:36:45

Let $A$ be symmetric positive definite. Given its eigendecomposition $$A=Q\Lambda Q^T$$ it's easy compute the determinant of $I + A$ as

$$I + A= QQ^T +Q\Lambda Q^T = Q(I+\Lambda)Q^T$$which is similar to $$I+A = \text{diag}(1+\lambda_1 ,\dots,1+\lambda_n)$$

Then the determinant is $$\vert I + A \vert= \prod_k (1+\lambda_k)$$

Now, suppose that instead of the eigendecomposition of $A$, we are given the Cholesky factorization $$A=LL^T$$

Is there still a computationally efficient method for calculating $\vert I + LL^T \vert$?