Given arbitrary real nonzero vectors $v$ and $u$, we define positive definite matrices $M=vv^T$ and $N=uu^T$ and their difference $W=M-N$. Extensive numerical calculations show that $W$ can never be positive semidefinite however small the norm $|u|$ is; the exception is when $u$ is parallel to $v$. How to prove this?
(I'm interested in the 3D case, but the conclusion seems to be independent of the dimension.)
Recall that a matrix $M$ is positive semidefinite iff $z^TMz \geq 0$ for all vectors $z$.
Let $z$ be an arbitrary vector which we'll choose in a second. Then for your matrix $M$, we have $$ \begin{align} z^TMz &= z^Tvv^Tz - z^Tuu^Tz \\ &= x^Tx - y^Ty \end{align} $$ where $x = v^Tz$ and $y=u^Tz$.
If $u$ and $v$ are not parallel, then we may pick $z$ perpendicular to $u$ and not $v$, forcing $y = 0$ and this quantity to be positive. But we could instead pick $z$ perpendicular to $v$ and not $u$, forcing $x = 0$ and this quantity to be negative.
Therefore, this difference is never positive semidefinite or negative semidefinite unless $u$ and $v$ are parallel, as you have observed numerically.