Assume I have a 3D vector $v$ and $\times$ is a dyadic product.
Is matrix
$$ M = v \times v' $$
always positive semi-definite?
Computing $x^T M x$ gives us $(x_1 v_1 + x_2 v_2 + x_3 v_3)^2$, which is always positive or zero. Have i missed anything in the proof?
Yes, for any vector $v \in \mathbb{R}^{n \times 1}$, $M=vv^{\intercal}$ is always semi-positive definite.
$\forall x \in \mathbb{R}^{n \times 1}$, $x^{\intercal}Mx=x^{\intercal}vv^{\intercal}x = x^{\intercal}v(x^{\intercal}v)^{\intercal} = (x^{\intercal}v)^2 \ge 0$. Note that $x^{\intercal}v$ is a scalar, thus $x^{\intercal}v=(x^{\intercal}v)^{\intercal}$.
However, $M$ may not be positive definite when $v \ne \mathbf{0}$ and $n > 1$: $x^{\intercal}Mx = 0 \Leftrightarrow v^{\intercal}x = 0$, where there is non-zero solution for $x$.