I will assume that we restrict our interest to real symmetric matrices $T$.
Both Positive Definite (PD) and Negative Definite (ND) matrices satisfy your definition of definiteness.
In the sequel, we show that these are the only two families of symmetric matrices that satisfy the definition. Equivalently, for any symmetric matrix $T$ that is neither PD, nor ND, there exists $x$ such that $\langle Tx,x \rangle=0$.
Assuming $T$ is neither PD nor ND, one of the following cases must be true:
- $T$ has an eigenvalue $\lambda_{i}=0$. In that case, $T$ is clearly not definite: for $x = u_{i}$, the eigenvector corresponding to $\lambda_{i}$, $\langle Tx,x \rangle=0$.
- $\lambda_{i} \neq 0$ $\forall i \in [n]$. In that case, $T$ is indefinite; it has both positive and negative eigenvalues. Without loss of generality, let $\lambda_{1}>0$, $\lambda_{n} < 0$, and let $u_{1}$, $u_{n}$ be the respective eigenvectors.
Let $x = \frac{1}{\sqrt{\lambda_{1}}}u_{1} +\frac{1}{\sqrt{|\lambda_{n}|}}u_{n}$.
Then,
$$
Tx = U\Lambda U^{T} x
=
\sqrt{\lambda_{1}}\cdot u_{1} - \sqrt{|\lambda_{n}|}\cdot u_{n},
$$
and
$$
\langle Tx, x\rangle
=
\langle \sqrt{\lambda_{1}}\cdot u_{1} - \sqrt{|\lambda_{n}|}\cdot u_{n}, \frac{1}{\sqrt{\lambda_{1}}}u_{1} +\frac{1}{\sqrt{|\lambda_{n}|}}u_{n}\rangle
=1 -1 = 0.
$$