TITLE: Does the following hold for any matrix $A$ with non-negative eigenvalues?
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: $$
A =
\left(
\begin{array}{cc}
1 & 9 \\
0 & 2
\end{array}
\right)
$$
and your
$$
x =
\left(
\begin{array}{c}
1  \\
-1
\end{array}
\right)
$$