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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.

Pablo Jeken Rico
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  • 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 Rico 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 Rico 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 Rico Sep 25 '19 at 17:47

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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.

flawr
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  • 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 Rico Sep 25 '19 at 17:41
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    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 Rico Sep 25 '19 at 19:10