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 $W'$.

  • $\begingroup$ Maybe $F$ should be an infinite field? I guess the statement is false otherwise. $\endgroup$ – Hugo Sep 27 '18 at 17:09
  • $\begingroup$ 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? $\endgroup$ – Matthew Leingang Sep 27 '18 at 17:15
  • $\begingroup$ I suppose you mean $\;W\color{red}\oplus W'=V$. $\endgroup$ – Bernard Sep 27 '18 at 19:01
  • $\begingroup$ @Bernard, no I did not mean direct sum since I am not requiring that $W \cap W' $ is trivial. $\endgroup$ – Vitolo Sep 27 '18 at 19:40
  • $\begingroup$ This is very unusual. In this case, $V$ itself is a complement of $W$. $\endgroup$ – 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 $W_1, \dotsc, W_l$.

  • $\begingroup$ Thanks @Hugo . Yes $F$ is infinite and I assumed that it was not needed. In fact, it is a crucial assumption. $\endgroup$ – Vitolo Sep 27 '18 at 18:19

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