I am reading a textbook which seems to imply that if $Q$ is a positive semidefinite matrix (not necesarily symmetric) and $x^TQx = 0$, then $x \in \text{Null}(Q).$ This seems a little suspicious to me, so to find a counterexample I tried to following: I tried looking for a vector $x$ and a $2 \times 2$ matrix $Q$ so that $$Q \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \begin{bmatrix} -x_2 \\x_1 \end{bmatrix}$$ If we find a non-zero $x$ and a p.s.d. $Q$ which satisfy the above equation, then we have a counterexample. However, I was not succesfull.
Is the proposition actually true, or is there a counterexample?
You can choose $Q$ as \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}
SHARE IDEA: How can we choose such matrix? Note that we can consider $Qx$ as an image of $x$ through a transformation. Now, we see that $$x^TQx=x \cdot (Qx)$$ Follow your idea, we aim to choose $Q$ such that the dot product between $x$ and its image is non-negative and equal to $0$ for some non-zero $x$. Thus, the transformation can be chosen as a $90^o$-rotation. So, we obtain $Q$ as above.