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The leading principal minors (Sylvester's criterion) is known for positive definite matrices.

However, I heard the following criterion to check whether a $\mathbf {real}$ square symmetric matrix $A$ is positive semi-definite:
$\Delta_1>0,\ldots,\Delta_{n-1}>0$ and $\det A=0$ iff $A\succeq 0$.

Is that true? I could not find a counter example nor prove it. The question rises because this criterion was known to me to check positive definiteness and never positive semi-definiteness, so I find it strage that such a simple criterion (if it is true) was unknown to me.

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Yes indeed Sylvester's criterion can be also used to determine the general signature of a matrix on the basis of the fact that the signature corresponds to the number of positive, negative and null eigenvalues.

For example this matrix

$$A=\begin{bmatrix}0&\frac12&0&0\\\frac12&0&0&0\\0&0&1&0\\0&0&0&0 \end{bmatrix}$$

has signature

$$n_+=2,n_-=1,n_0=1$$

indeed since $\det(A)=0$ at lest one eigenvalue is equal to $0$, then consider

$$B=\begin{bmatrix}0&\frac12&0\\\frac12&0&0\\0&0&1 \end{bmatrix}$$

and $\det(1)>0$ thus at least one eigen value is positive and since $\det(B)<0$ the only possible signature for $B$ is $$n_+=2,n_-=1$$

What about this one?

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -5\\ 0 & 0& 0 & -5 & 0\\ 0 & 0 & -5 & 0 & 0\\ 0 & -5 & 0 & 0 & 0\\ \end{bmatrix}$$

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  • $\begingroup$ I do not understand why this means that $\Delta_1>,\ldots ,\Delta_{n-1}>0,\det A=0 \Rightarrow A\succeq 0$. Am I missing something trivial? $\endgroup$ – MasterJ Apr 19 '18 at 13:35
  • $\begingroup$ @MasterJ recall that $\det(A)=\prod \lambda_i$ $\endgroup$ – gimusi Apr 19 '18 at 13:36
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    $\begingroup$ Of course. Your answer made things clearer. Thanks. $\endgroup$ – MasterJ Apr 19 '18 at 14:05
  • $\begingroup$ Well done! You are welcome! Bye $\endgroup$ – gimusi Apr 19 '18 at 14:48

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