Let $W$ be a skew-symmetric matrix of dimension $n$ with $\det(W)=0$. If $1\leq \text{Null}(W)<n$, it's clear that the image has dimension greater that 1. My question is if $\beta$ is a basis of image of $W$, could it happen that $\beta\cap \ker(W)\ne \emptyset$ or $\beta\cap \ker(W)\ne \left\{0\right\}$?
In fact, I only need to know if $Wv\ne 0$ for all elements in $\beta$.
I really need help with this problem.
Assuming real matrices, you have the general equality $$ \ker W = (\text{ran}\,W^T)^\perp. $$ Since $W^T=-W$ and $\text{ran}\,(-W)=\text{ran}\,W$, you get $$ \ker W = (\text{ran}\,W)^\perp. $$ So $$\ker W\,\cap\,\text{ran}\,W=\{0\}.$$