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Let's call a matrix $M \in R^{n \times n}$ strictly positive definite if for any $c \in R^n$, $c \neq 0$, we have $c^T M c > 0$. Consider arbitrary matrix $Q \in R^{n \times n}$ that is not strictly positive definite. Is it true that we can find $c \in R^n$, $c \neq 0$, such that $c^T Q c = 0$?

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Let $S:=\{x\in\Bbb R^n : \sum_{j=1}^nx_j^2=1\}$. The map $x\in S\mapsto x^tQx$ is continuous (as polynomial in $x_j$, $1\leq j\leq n$, so its minimum $m$ on $S$ is reached, as it's a compact set. So is the maximum $M$.

If $m\leq 0\leq M$, it's OK, as the connected set $S$ is mapped to an interval, if $M<0$ or $m>0$ we can't find $c\neq 0$ such that $c^tQc=0$.

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(+1) Great answer, thank you. Just a little expansion. $x^t Q x \geq m > 0$ if and only if $Q$ is strictly positive definite. $x^t Q x \leq M < 0$ if and only if $Q$ is strictly negative definite. $m \leq 0 \leq M$ means that $Q$ is not strictly positive nor strictly negative definite. In this case it has been shown that there exists $c \neq 0$: $c^t Q c = 0$. –  Leo Oct 28 '12 at 15:46
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