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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.

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  • $\begingroup$ 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. $\endgroup$ Sep 25, 2019 at 16:19
  • $\begingroup$ 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. $\endgroup$
    – user
    Sep 25, 2019 at 16:30
  • $\begingroup$ The $0$ is missing right? It could be semidefinite... $\endgroup$ Sep 25, 2019 at 16:36
  • $\begingroup$ If "A is not positive definite" of course it can also have only one negative eigenvalue. $\endgroup$
    – user
    Sep 25, 2019 at 16:47
  • $\begingroup$ 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 $det(A)<0$. $\endgroup$ Sep 25, 2019 at 17:47

1 Answer 1

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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.

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  • $\begingroup$ 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). $\endgroup$ Sep 25, 2019 at 17:41
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    $\begingroup$ 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. $\endgroup$
    – flawr
    Sep 25, 2019 at 18:33
  • $\begingroup$ That makes a lot of sense, thank you very much!! $\endgroup$ Sep 25, 2019 at 19:10

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