Is there a theorem or a way to show that if I have a real and symmetric positive definite matrix $A$ and its Cholesky factorization is $A = LL^T$ then $B = L^TL$ is also positive definite? Or in other words, the Cholesky factorization of $B$ also exists?
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$\begingroup$ Yes, if $X$ is square nonsingular, $X^TX$ and $XX^T$ are positive definite. $\endgroup$– Algebraic PavelApr 18, 2015 at 11:32
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1 Answer
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First, note that the inverse of a lower triangular invertible matrix (such as $L$ from the Cholesky decomposition) is also lower triangular. So, $R=L^{-1}$ is lower triangular (and invertible). Thus, $R R^T$ is the Cholesky decomposition of some positive definite matrix. Inverting $R R^T$ gives $(R^{-1})^T R^{-1} = L^T L$. Since $R R^T$ was positive definite, so is its inverse, so $L^T L$ is positive definite as well.
