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This is an intermediate step in a probability homework problem. I have all of it done except for this one step of justification which (I hope!) is true.

Let $\Sigma$ be the covariance matrix of an $n$-dimensional Gaussian. Suppose $\forall i,j,\Sigma_{i,j}>0$. By properties of covariance, $\Sigma$ is symmetric. Show $\Sigma$ is positive definite.

I'm unfortunately getting stuck on the algebra, which ironically probably should be the easiest part of the problem. It's easy enough to show that $\Sigma$ is positive semidefinite (I used the same method here), but I'm having a hard time ruling out the possibility that there is some nonzero $x\in\mathbb R^n$ such that $x^\mathrm T\Sigma x=0$.

Am I'm missing something obvious?

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  • It's not clear what the title of your question has to do with the question itself – Ben Grossmann Feb 03 '17 at 00:22
  • I have a symmetric matrix with (strictly) positive entries. I'm trying to show it's (strictly) positive definite. The quoted section of the question should match the title. – user42869 Feb 03 '17 at 00:23
  • The problem is that a symmetric matrix with strictly positive entries is not necessarily positive-semidefinite. So it is not clear (to me, at least) what you are asking. – Martin Argerami Feb 03 '17 at 00:24
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    Well, positive entries won't let you deduce that the matrix is positive definite. For example, take $$\pmatrix{1&1\1&1}$$ – Ben Grossmann Feb 03 '17 at 00:25
  • Or, worse, $$\begin{bmatrix} 1&2\ 2&1\end{bmatrix}$$ is not even positive-semidefinite. – Martin Argerami Feb 03 '17 at 00:26
  • Yes, you're right. The matrix explicitly has to be a covariance matrix. Lemme edit the question. – user42869 Feb 03 '17 at 00:26
  • Indeed :D $\ \ \ \ $ – Martin Argerami Feb 03 '17 at 00:27
  • @user42869 any positive semidefinite matrix can be a covariance matrix – Ben Grossmann Feb 03 '17 at 00:30
  • Yes, but I'm asking if a covariance matrix with strictly positive entries is positive definite. – user42869 Feb 03 '17 at 00:31
  • Perhaps you can show $\Sigma$ is inveritble? This would be sufficient. – Aweygan Feb 03 '17 at 00:32
  • To be honest, it's not obvious to me why that would be sufficient. – user42869 Feb 03 '17 at 00:33
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    @user42869 we just told you that $\pmatrix{1&1\1&1}$ is a positive semidefinite matrix with strictly positive entries. In addition, because it is positive semidefinite, it is also a covariance matrix (of some distribution). So no, your claim does not hold. – Ben Grossmann Feb 03 '17 at 00:34
  • @Omnomnomnom I see what you mean. OK, that answers my question. Thanks! – user42869 Feb 03 '17 at 00:35
  • @user42869 it seems to me that you can use the steps in the linked post to show that $x^T\Sigma x > 0$ for any $x$. – Ben Grossmann Feb 03 '17 at 00:43
  • I only see the proof for positive semi-definiteness, but your correction of my assumptions caused me to go back and realize that I didn't actually need positive definiteness for my proof.

    I only needed it to be positive semi-definite, which (as noted in the link) is straightforward to show. Dumb mistake.

     – user42869 Feb 03 '17 at 00:46

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