The complement is not uniquely determined (although it's dimension is). Consider the complement of the subspace $U$ spanned by $\begin{pmatrix}1 & 0\end{pmatrix}^\mathrm{T}$ in $\mathbb{R}^2$. Any subspace spanned by any vector which is not a multiple of $\begin{pmatrix}1 & 0\end{pmatrix}^\mathrm{T}$ will serve as a valid complement.
This complement differs from the set theoretic complement in a major way. $V\setminus U$ is a set but not a vector space in general; it simply removes the portion of $W$ in $V$. In the above example, $\mathbb{R}^2\setminus U$ is simply the $xy$-plane with the $x$-axis carved out. It does not have the closed structure of a vector space.
A more natural analogue of the set-theoretic complement is given by $V/U$ which is a quotient space. In this structure, we not only remove the vectors of $U$ but also compress the components of the remaining vectors in the direction of $U$ to retain a vector space structure.