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Prove if the determinant of a matrix is positive, then it has a Cholesky factorization.

A cholesky factorization can only be performed for hermitian, positive definite matrices. Should I go about proving that if the determinant of a matrix is zero, then that matrix is positive definite&hermitian? This seems circuitous. Anyone have other suggestions?

roulette01
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    It is wrong, $\begin{bmatrix} -1 & \ & -1 \end{bmatrix}$ doesn't have a cholesky decomposition. – user251257 Mar 29 '16 at 01:33
  • I think I am also supposed to assume the matrix is hermitian and positive definite. How would you go about proving in this case? – roulette01 Mar 29 '16 at 01:45
  • than it is true. Hermitian and positive definite implies positive determinant. The converse is wrong. – user251257 Mar 29 '16 at 01:46

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