Does the following hold for any matrix $A$ with non-negative eigenvalues?

If $A\in\Re^{q\times{q}}$ is a square matrix with non-negative eigenvalues. Is it possible to show that $x^TAx\geq{0}$ for any non-zero vector $x$? I know this is obvious for positive semi-definte matrices but when it is not, is there a way to show that $x^TAx\geq{0}$?

I know this holds for an eigenvector $x$ corresponding to eigenvalue $\lambda$, $Ax=\lambda{x}$ and $x^T{A}x=\lambda\|x\|^2\geq{0}$. But in case when $x$ is not an eigenvector is that claim true?

$$A = \left( \begin{array}{cc} 1 & 9 \\ 0 & 2 \end{array} \right)$$ and your $$x = \left( \begin{array}{c} 1 \\ -1 \end{array} \right)$$