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I can get why a matrix being positive definite implies all it's leading submatrices have positive determinant. However I'm unable to prove the converse and I couldn't get all the answer to this converse implication I have come across on this substack. All of them uses theorems that I'm completely unaware of or just couldn't prove it.

The text on Linear Algebra that I'm following i.e Linear Algebra by Friedberg, Incel & Spence and a book written by Gilbert Strang have very little literature on positive definite matrix. The former just included the positive definite matrix as a question in one of it's exercises. Strang's book just assume that you already know about positive definite matrix.

Which book I could read to learn more about positive definite matrix in detail?

Minto P
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    There is a proof here: https://en.wikipedia.org/wiki/Sylvester%27s_criterion – Kavi Rama Murthy Oct 22 '24 at 05:41
  • Possibly helpful: https://math.stackexchange.com/q/811387/42969 – Martin R Oct 22 '24 at 05:44
  • @MartinR The top answer is what I first came across for the converse proof. Problem is, I couldn't get how they could assert that a non-positive Hermitian matrix would have two mutually orthogonal eigenvectors with negative eigenvalues. Otherwise, the proof is sound. – Minto P Oct 22 '24 at 06:23
  • @MintoP Where are you getting lost? Perhaps this explanation will be helpful: it's a proof by contradiction, so we're assuming that $A$ has positive leading minors but is not positive definite. Because the determinant is $0$ but has some non-positive eigenvalues, $A$ has at least 2 negative eigenvalues (up to multiplicity). By the spectral theorem, two mutually orthogonal eigenvectors can be found associated with these eigenvalues. – Ben Grossmann Oct 22 '24 at 15:07
  • @BenGrossmann your explanation doesn't make any sense to me. 1) How can you say that the determinant is zero? It being positive definite doesn't mean that it'd have zero determinant. It may have a negative determinant if one of it's eigenvalue is negative. 2) How can you be so sure that it'll have at least 2 negative eigenvalues? We can't even be sure if it has a negative eigenvalues at all. – Minto P Oct 24 '24 at 04:44
  • @MintoP First of all, I meant to say that the determinant is not $0$, my mistake – Ben Grossmann Oct 24 '24 at 14:41
  • @MintoP Second, we have assumed (for the purpose of contradiction) that the Hermitian matrix $A$ (which has positive leading principal minors) is not positive definite. The fact that the determinant is positive means that the product of its eigenvalues (which are real) is a positive number. In order for the product of real numbers to be positive, an even number of these numbers must be negative. Since we know that $A$ is not positive definite, its eigenvalues cannot all be positive, so we conclude that at least two of its eigenvalues are negative. – Ben Grossmann Oct 24 '24 at 14:44

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