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Let $A$ be an $n\times n$ symmetric matrix. Then, $A$ is a positive semidefinite iff every principal minor of $A$ is $\geq0$; $A$ is a negative semidefinite iff every principal minor of odd order is $\leq0$ and every principal minor of even order is $\geq0$.

Now, let $B$ be a $4\times 4$ symmetric matrix with

  • principal minor of odd order $=0$ and principal minor of even order $\geq0$.
  • all principal minors $=0.$

I think in both the cases we have positive as well as negative semidefinite matrices, but I think how a single matrix be both +ve and -ve?

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    $\begingroup$ $B$ also symmetric? $\endgroup$ – Yurii Savchuk Dec 6 '13 at 12:45
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    $\begingroup$ If all principal minors of $B=B^*$ are $0$ then $B=0.$ $\endgroup$ – Yurii Savchuk Dec 6 '13 at 12:48
  • $\begingroup$ @YuriiSavchuk, thanks for poing error out. What is $B^*$ in second comment? $\endgroup$ – Silent Dec 6 '13 at 12:54
  • $\begingroup$ $B^*=B^T$ over $\mathbb R$ $\endgroup$ – Yurii Savchuk Dec 6 '13 at 12:55
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For all principal minors $=0$, you got your answer in Yurii's comment: such matrix is equal to the zero matrix, which is both positive and negative semidefinite.

As for the case

principal minor of odd order $=0$ and principal minor of even order $\ge 0$,

consider all principal minors of order $1$. They are equal to zero, which means that the diagonal elements of $B$ are equal to zero.

Now consider all principal minors of order $2$: they are $\ge 0$. But, these are determinants of symmetric matrices of order $2$ with zero diagonal, so

$$0 \le \det \begin{bmatrix} 0 & b_{ij} \\ b_{ij} & 0 \end{bmatrix} = -b_{ij}^2,$$

so $b_{ij} = 0$ for all $i \ne j$. In other words, $B = 0$, as in your other case.

Even without this analysis, the answer is quite obvious: by your condition, $B$ is both positive and negative semidefinite. It is easy to show that this is possible if and only if $B = 0$.

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  • $\begingroup$ Thank you so much! Never expected such informative answer to seemingly so silly question! Thanks! $\endgroup$ – Silent Dec 6 '13 at 14:58
  • $\begingroup$ You're welcome. If the question does not warrant closing, then it's worth properly answering. :-) $\endgroup$ – Vedran Šego Dec 6 '13 at 15:00

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