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Choelsky Decomposition allows us to decompose a Hermitian Positive Definite Matrix $A$ as $A=LL^*$, and $L$ is guaranteed to be lower-triangular.

My Question: If $A$ is non-Hermitian, but still positive definite, can we still find a matrix $B$ such that $A=B^*B$ (assuming nothing about $B$)? If not, are there any looser constraints on $A$?

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A similar (to user1551's) argument applies for using only transpose (as opposed to the transpose conjugate). If $A=B^\top B$ then $$A^\top = (B^\top B)^\top = B^\top B = A$$

This shows that $A$ must be symmetric. If it has complex values then it would not be Hermitian as in that case it must be conjugate transpose, not just transpose.

However, I have found it relatively simple to find such a similarity, so that a symmetric (not Hermitian) matrix is similar to $A$: $$A \sim QAQ^{-1} = Q^{-\top}A^\top Q^\top$$

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If $A=B^\ast B$, the matrix $A$ must be Hermitian and positive semidefinite: $A^\ast=(B^\ast B)^\ast = B^\ast B=A$ and $x^\ast Ax=x^\ast B^\ast Bx=\|Bx\|^2\ge0$. So, if $A$ is non-Hermitian, you cannot decompose it into a product of the form $B^\ast B$.

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Of course. Sorry I missed that. I meant to loosen only the symmetry constraint, and went a bt overboard... I'm editing my question. –  yohBS Feb 3 '13 at 18:42
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