# Intersection of subspaces equals to another subspace

I'm having problems with proving the following proposition:

Let $W$ be subspace of an n-dimensional vector space $V$($W$ is considered to have dimension $r<n$). Show that

$W=\bigcap_{U\subseteq V\text{ is linear, }dim(U)=n-1,W\subseteq U}U$

I.e., show that $W$ is equal to the intersection of all n-1-dimensional subspaces containing $W$.

My approach was the following: obviously $W$ is contained in the intersection and thus a subspace of it. Thus for showing that both are equal it suffices to show that their dimensions are equal, i.e. that the dimension of the intersection is $r$. Trying out variations of the dimension formula, I'm now stuck with this.

If you assume that there is some vector outside of $W$ but in the interection of all those subspaces, you can use this vector to construct a codimension $1$ subspace containing $W$ for a quick contradiction.

• How exactly would this look, would I extend the single vector with W to another subspace?
– blub
Nov 30, 2017 at 15:34
• You want this vector to not be in the intersection of the subspaces, or in particular you want to construct a subspace containing W but not containing this vector, with dimension n-1. Nov 30, 2017 at 15:38
• How would I construct this space?
– blub
Nov 30, 2017 at 22:07
• Okay, my hint doesnt seem to be helping so ill just tell you how to do it. Suppose $e_1$ is a vector not in $W$. It's non zero as its not in $W$, so extend it to a basis of $V$: $e_1, \ldots, e_n$. Consider the subspace $U:=<e_2, \ldots, e_n>$. Its easy to show this has dimension $n-1$ and contains $W$, so $e_1$ is not in the right hand side. I hope this has cleared things up. Dec 1, 2017 at 15:39