If matrix is not positive semidefinite then there is $x$ such that $x^T A x < 0$

Question

Is it true that if symmetric matrix $A \in \Bbb R^{n \times n}$ is not positive semidefinite then there is $x \in \Bbb R^n$ such that $x^T A x < 0$? Because in this case it can be indefinite but why is there necessarily some $x$ for which $x^T A x < 0$?

Answer 1

This is simply by negating the definition: A positive semi-definite matrix $A$ is defined as a matrix, such that $$\forall \ x: \ x^TAx \geq 0. $$ The negation of this is:$$ \exists \, x: x^TAx<0. $$