# Complement of finitely many vector subspaces

Let $$F$$ be a field and $$V$$ be a $$n$$-dimensional vector space over $$F$$.

For any subspace $$W\subseteq V$$ we say that $$W'$$ is a complement of $$W$$ if $$W + W' = V.$$

I have the following question. Consider $$W_1,\ldots, W_l$$ finitely many subspaces of the same dimension $$k$$, then prove that there exists a subspace $$W'$$ of dimension $$n-k$$ such that it is a complement for all $$W_i$$.

I have tried by considering basis $$B_i$$ of $$W_i$$ and extending to basis of $$V$$; but I do not really know how to "simultaneously" extend the basis so I can define the complement $$W'$$.

• Maybe $F$ should be an infinite field? I guess the statement is false otherwise. – Hugo Sep 27 '18 at 17:09
• Have you tried a few examples. For instance, $V = \mathbb{R}^2$, and $W_1$ and $W_2$ the subspaces generated by $e_1$ and $e_2$? Then $W'$ can be the subspace generated by $e_1 + e_2$. What if $V = \mathbb{R}^3$ with the same two subspaces? – Matthew Leingang Sep 27 '18 at 17:15
• I suppose you mean $\;W\color{red}\oplus W'=V$. – Bernard Sep 27 '18 at 19:01
• @Bernard, no I did not mean direct sum since I am not requiring that $W \cap W'$ is trivial. – Vitolo Sep 27 '18 at 19:40
• This is very unusual. In this case, $V$ itself is a complement of $W$. – Bernard Sep 27 '18 at 19:52

I assume that $$F$$ is an infinite field, oherwise the statement is false.
By (backward) induction on $$k$$. If $$k = n-1$$, then pick any vector $$v$$ such that $$v \notin W_j$$ for every $$j$$ and let $$W' = \mathrm{Span}(v)$$.
Given $$W_1, \dotsc, W_l$$ of dimension $$k$$, take any vector $$v$$ such that $$v \notin W_j$$ for every $$j$$, and let $$W'_j = \mathrm{Span}(W_j, v)$$. Then $$\dim W'_j = k+1$$, and we apply the induction hypothesis on $$W'_j$$ to get $$W'$$ of dimension $$n-k-1$$ a complement for all $$W'_j$$. In particular this implies that $$v \notin W'$$, hence $$\mathrm{Span}(W', v)$$ is of dimension $$n-k$$ and is a complement for $$W_1, \dotsc, W_l$$.
• Thanks @Hugo . Yes $F$ is infinite and I assumed that it was not needed. In fact, it is a crucial assumption. – Vitolo Sep 27 '18 at 18:19