Unique complement of a subspace Can we have a unique complement to a subspace of a finite-dimensional vector space $V$ (exclude the trivial cases $0$ and $V$)? 
 A: If $V$ only has the structure of a vector space, then given a proper, nontrivial subspace $W\subset V$, there is no canonical way of choosing a subspace $U$ such that $W\oplus U=V$.
Just think of a line through the origin in $\mathbb{R}^2$.  Any different line through the origin is a complement of the original line.  In general, if $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then any $(n-m)$-dimensional subspace $U$ that intersect $W$ trivially is a complement.
However, if we place a bilinear form on $V$, then we can talk about orthogonal complements, which are canonical.  In the case of our example, there is a distinguished complement to our original line: the line that is perpendicular.  This is the orthogonal complement that arises from the usual dot product on $\mathbb{R}^2$.
A: No. Let $U$ be your subspace, with $\{ 0 \} \subset U \subset V$. (I mean, proper inclusions.) Let $W$ be a complement of $U$ in $V$, and pick an element $0 \ne w \in W$. Let $B$ be a basis of $W$ including $w$. Pick $0 \ne u \in U$, and modify $B$ by replacing $w$ with $w + u$, to obtain a new set $B'$. Now $W' = \langle B' \rangle$ is another complement of $U$, different from  $W$.
