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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 – 2017-02-03
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    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. – 2017-02-03
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    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. – 2017-02-03
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    Well, positive entries won't let you deduce that the matrix is positive definite. For example, take $$\pmatrix{1&1\\1&1}$$ – 2017-02-03
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    Or, worse, $$\begin{bmatrix} 1&2\\ 2&1\end{bmatrix}$$ is not even positive-semidefinite. – 2017-02-03
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    Yes, you're right. The matrix explicitly has to be a covariance matrix. Lemme edit the question. – 2017-02-03
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    Indeed :D $\ \ \ \ $ – 2017-02-03
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    @user42869 any positive semidefinite matrix can be a covariance matrix – 2017-02-03
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    Yes, but I'm asking if a covariance matrix with strictly positive entries is positive definite. – 2017-02-03
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    Perhaps you can show $\Sigma$ is inveritble? This would be sufficient. – 2017-02-03
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    To be honest, it's not obvious to me why that would be sufficient. – 2017-02-03
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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. – 2017-02-03
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    @Omnomnomnom I see what you mean. OK, that answers my question. Thanks! – 2017-02-03
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    @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 $x$. – 2017-02-03
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    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. – 2017-02-03

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