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I would like to know how to prove that the inverse of a covariance matrix $\Sigma^{-1}=\Omega$ is positive semi-definite too.

My second question can we prove that for any matrix $A\in\cal{M}_{nm}$ we have the determinant $|A^T~\Omega~A|\ge0$.

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Hint. Every covariance matrix is positive semidefinite. Hence every invertible covariance matrix is positive definite. Now, every positive definite matrix is unitarily diagonalisable. Therefore ...

For your second question, presumably the matrices are real. Consider the quantity $x^TA^T\Omega Ax$. Do you see that it is always nonnegative?

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  • $\begingroup$ Actually, I am a grad student and my last course in algebra was in 2008. Anyways, I could answer this by considering the eignevalues of the inverse which are positive hence the positive definiteness.. Regarding the second question I think it's easy to prove that by doing the cholesky decomposition of $\Omega$ then we can prove that the prodcut is positive definite since it's of the form $(\cdot)^T(\cdot)$ which is nothing but the squared $L_2$ norm that is always positive.. $\endgroup$ – user2987 Apr 16 '14 at 19:46
  • $\begingroup$ @Bel Yes, that's a nice way to answer your question. $\endgroup$ – user1551 Apr 16 '14 at 20:00

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