Matrices which never map vectors onto orthogonal vectors Clearly, there are matrices $M$ for which there exists a vector $v$ such that $\langle v,Mv \rangle=0$ but $M v\ne 0$. In fact, a $90^{\circ}$ rotation in $2D$ has this property for all $v$:
$$v^T \left[\begin{matrix}0 & 1 \\ -1 & 0\end{matrix}\right] v \equiv 0$$
What can we say about matrices that don't have this property? Matrices for which $\langle v,Mv \rangle=0 \implies M v=0$. The unit matrix is an example, as is
$$\left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right] $$
so there are non-invertible examples.
How can we check that $\langle v,Mv\rangle=0\implies M v=0$? Does it suffice that $M$ is symmetric/Hermitian?
 A: We have $\left<v,Mv\right>=\frac12\left<v,(M+M^t)v\right>$
where $M^t$ is the transpose of $M$. Then $M_1=M+M^t$ is symmetric.
If the matrix $M_1$ is definite (positive or negative)
then $\left<v,Mv\right>=0$ implies $v=0$. If $M_1$ is indefinite,
but non-singular, there will be $v$ with $M_1v\ne0$ and $\left<v,Mv\right>=0$.
A: The matrices that can create zero in this way are precisely the indefinite matrices. 
It follows directly from the definitions that positive or negative definite matrices cannot create zeros in this way.
For indefinite matrices, we can always create such a zero by solving a quadratic equation. Suppose 
\begin{align}
c :=&v^* M v>0 \\
a:=&u^*M u<0\\
b:=&u^*(M+M^*)v \in \mathbb{C}.
\end{align}
Then plugging the linear combination $v+zu$ for complex weight $z=r e^{i\theta}$ yields:
$$(v+zu)^* M (v+zu) = a r^2 + (b e^{i\theta}) r + c.$$
We can choose the phase $\theta$ so that 
$$b_2 :=b e^{i\theta}>0$$
is a positive real number, and choose the magnitude $r$ as the real positive solution to the associated quadratic equation. This is always possible since the discriminant,
$$b_2^2 - 4 a c >0$$
Is always positive and larger than $b_2^2$ (recall $a$ is negative and $c$ is positive).
Hence whenever the matrix is indefinite, we can always find a linear combination so that $(v+zu)^* M (v+zu)=0$.
