Eigenvalues of a positive principal minors symmetric matrix

I am trying to understand the proof of the Sylvesters Criterion. More concretely:

Suppose that the real symmetric matrix $$A$$ has only positive principal minors.

The statement that I do not understand says: "It follows that if $$A$$ is not positive definite, it must possess at least two negative eigenvalues.". Why can't $$A$$ just have one negative eigenvalue? I would be very grateful if somebody could explain this or give me a good source to read through.

• Why is that? I am trying to prove that it cannot be non positiv definite. And to my knowledge a positive definite matrix has only positive eigenvalues. – Pablo Jeken Sep 25 '19 at 16:19
• Sorry I mean that the statement you are reffering to "It follows that if A is not positive definite, it must possess at least two negative eigenvalues" is uncorrect. – user Sep 25 '19 at 16:30
• The $0$ is missing right? It could be semidefinite... – Pablo Jeken Sep 25 '19 at 16:36
• If "A is not positive definite" of course it can also have only one negative eigenvalue. – user Sep 25 '19 at 16:47
• Well, as @flawr has shown for $det(A)>0$ it is clear that a non positive definite matrix has to have at least two negative eigenvalues. Moreover it can just have an even amount of negative eigenvalues, otherwise $det(A)<0$. – Pablo Jeken Sep 25 '19 at 17:47

We know that $$\det(A) > 0$$. Recall that $$\det(A) = \prod_i \lambda_i$$ where $$\lambda_i$$ are the eigenvalues of $$A$$. Then we cannot just have one negative eigenvalue if $$A$$ has a positive determinant.

• Thanks! I stumbled upon a further small question regarding the same proof. By showing that the last entry of a vector is not accountable for a negative result of $v^T Av$, the author states: "Hence the leading $(n−1)×(n−1)$ principal submatrix of A is not positive definitive." Why is that so? (author is inacive). – Pablo Jeken Sep 25 '19 at 17:41
• Let us write $A = \begin{bmatrix} A' & a \\ a^* & a_{nn} \end{bmatrix}$ where $A'$ is th leading principal $(n-1)\times(n-1)$ submatrix and let us write $u = \begin{bmatrix} w \\ 0 \end{bmatrix}.$ Then $0 > u^* A u = u^* \begin{bmatrix} A' w + 0a \\ a^* w + 0a_{nn}\end{bmatrix} = w^* (A' a +0a) + 0(a^*w + 0a_{nn}) = w^*A'w$, therefore $A'$ is not positive definite. – flawr Sep 25 '19 at 18:33
• That makes a lot of sense, thank you very much!! – Pablo Jeken Sep 25 '19 at 19:10