Given a real symmetric matrix $S$ of order $n$, assuming $\mathbf{x} \equiv \mathbf{e}_i$ with $i \in [1,\,n]$ we have:
$$ \mathbf{x}^t\,S\,\mathbf{x} = s_{i,i} $$
so if $s_{i,i}=0$ for at least one $i \in [1,\,n]$ it follows that $S$ cannot be definite (positive or negative).
If in addition $\det(S) \ne 0$, then all the eigenvalues of $S$ aren't zero and consequently $S$ is indefinite.
In this way we have proof a first thesis:
$$ \color{blue}{\boxed{\text{If a non-singular real symmetric matrix has a zero on the diagonal it's indefinite.}}} $$
On the other hand, after some numerical experiments, I noticed that $\det(S)=0$ implies that $S$ is a semi-definite matrix (positive or negative) only when the i-th row considered is all zeros.
This leads me to formulate a second thesis:
$$ \color{green}{\boxed{\text{If a real symmetric matrix has a non-zero row with a zero on the diagonal it's indefinite.}}} $$
For $n = 1$ or $n = 2$ the thesis is trivially true, while for $n = 3$ I thought I'd consider:
$$ S = \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & \color{\red}{0} \\ \end{bmatrix} $$
under the condition $c^2+e^2\ne 0$ and we start with the calculation of the determinant:
$$ \det(S) = 2\,b\,c\,e - a\,e^2 - c^2\,d\,. $$
Having done this, it's evident that if $c=0$ the determinant vanishes only if $a=0$ and in this case:
$$ \text{eig}(S) = \left\{0,\,\frac{d\pm\sqrt{d^2+4\left(b^2+e^2\right)}}{2}\right\} $$
where the two non-zero eigenvalues are always discordant and therefore $S$ is indefinite.
On the other hand, if $c \ne 0$ the determinant vanishes only if $d = \frac{2\,b\,c\,e-a\,e^2}{c^2}$ and in this other case:
$$ \text{eig}(S) = \left\{0,\,\frac{a\,c^2+\left(2\,b\,c-a\,e\right)e \pm \sqrt{c^2+e^2}\sqrt{a^2\,c^2+\left(2\,b\,c-a\,e\right)^2+4\,c^4}}{2\,c^2}\right\} $$
where the two non-zero eigenvalues are always discordant and therefore $S$ is indefinite.
For $n > 3$ the calculations get too complicated, ideas on how to proceed? Thanks!
