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Please note: I'm not interested in the difference between positive definiteness and semi-definiteness for this question.

A correlation matrix is a symmetric positive semi-definite matrix with 1s down the diagonal and off-diagonal terms $ -1 \leq M_{ij} \leq 1$.

Since a correlation matrix must be positive semi-definite, it must have a positive (or zero) determinant, but does a positive determinant imply positive definiteness? In other words, if I have a matrix with 1s down the diagonal, off-diagonals satisfying $ -1 \leq M_{ij} \leq 1$ and positive determinant, is that enough to prove that the matrix is positive definite (and thus an acceptable correlation matrix)?

Thank you.

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    $\begingroup$ $$\begin{bmatrix} -1 & 0\\ 0 & -1\end{bmatrix}$$has a positive determinant, but it's negative definite. $\endgroup$ – Rodrigo de Azevedo Jun 21 '16 at 10:38
  • $\begingroup$ Thanks, but this doesn't satisfy the set-up of my question - it has negative 1s along the diagonal, so clearly isn't a correlation matrix from the get-go. $\endgroup$ – JamesF Jun 21 '16 at 11:04
  • $\begingroup$ You asked "does a positive determinant imply positive definiteness?". $\endgroup$ – Rodrigo de Azevedo Jun 21 '16 at 11:04
  • $\begingroup$ Only in the first part! I clarified what I meant in the second sentence of the second paragraph. $\endgroup$ – JamesF Jun 21 '16 at 11:40
  • $\begingroup$ Your question is very messy. Correlation matrices are symmetric and positive semidefinite. Hence, their determinants are always nonnegative, but not necessarily positive. $\endgroup$ – Rodrigo de Azevedo Jun 21 '16 at 11:44
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The answer to your question, as I now understand it, is no. In particular, we can construct a matrix of you particular pattern with a positive determinant that fails to be positive definite.

In particular, consider the matrix

$$ M = \pmatrix{ 1&-1&-1&0&0&0\\ -1&1&-1&0&0&0\\ -1&-1&1&0&0&0\\ 0&0&0&1&-1&-1\\ 0&0&0&-1&1&-1\\ 0&0&0&-1&-1&1\\ } $$ which has eigenvalues $-1-1,2,2,2,2$

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  • $\begingroup$ Thanks. Could I ask you to expand to demonstrate this, or could you include a link that would do so? $\endgroup$ – JamesF Jun 21 '16 at 12:28
  • $\begingroup$ See my latest edit. $\endgroup$ – Omnomnomnom Jun 21 '16 at 12:37
  • $\begingroup$ Thanks, but I don't think this really answers my question. I'm not interested in the "semi-" part of "semi-definite". What I want to know is this: I have a matrix which looks like a correlation matrix (1s on the diagonal, 0.5 etc on the off-diagonal) and I know it has a positive determinant. Can I be sure from this, without calculating eigenvalues, that the matrix is PSD? ie. can I be sure that there aren't two negative eigenvalues lurking in there that would give it a positive determinant but still fail to be be positive semi-definite? $\endgroup$ – JamesF Jun 21 '16 at 13:18
  • $\begingroup$ See my latest edit... again $\endgroup$ – Omnomnomnom Jun 21 '16 at 13:36
  • $\begingroup$ Thanks! (side note: the matrix actually has eigenvalues of -1, -1, 2, 2, 2, 2) $\endgroup$ – JamesF Jun 21 '16 at 13:56
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Three facts:

  1. Symmetric semi positive definite matrix has all eigenvalues real and greater or equal zero.
  2. Symmetric positive definite matrix has all eigenvalues real and greater than zero.
  3. Determinant of any matrix is equal to product of all eigenvalues.

Hence, if determinant of symmetric semi positive definite matrix $A$ is nonzero, then $A$ is positive definite.

The answer is yes.

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  • $\begingroup$ Thanks, but like the other commentators you've focussed on determining semi-definite vs definiteness. That's not what I was asking - see Omnomnomnom's answer. $\endgroup$ – JamesF Jun 23 '16 at 6:55
  • $\begingroup$ Question: Since a correlation matrix must be positive semi-definite, it must have a positive (or zero) determinant, but does a positive determinant imply positive definiteness?. I understand this question as: is correlation matrix with positive determinant a positive definite matrix. My answer shows that this is true. $\endgroup$ – Pawel Kowal Jun 23 '16 at 7:43
  • $\begingroup$ Please read the very next sentence after the part of the question you've quoted. Better yet, read Omnomnomnom's answer. $\endgroup$ – JamesF Jun 24 '16 at 14:50
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Slight variation to original poster's question, would the conjecture hold if the determinants of leading principal minors are non-negative? I.e.

Take a matrix with the following properties: a) real symmetric b) all entries between -1 and 1 c) diagonal elements are all equal to 1 d) determinants of leading principal minors are non-negative

Is there an example of such a matrix which is not positive semi-definite? I wasn't able to construct any such example yet.

This is very much related to Sylvester's criterion and already holds for positive definite matrices, whereas in the case of positive semi-definite matrices, all principal minors must have non-negative determinants. The conjecture (which seems what the original poster was after) really means that for such "nicely" structured matrices, the stronger property holds and we only have to check the leading principal minors?

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