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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.

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1 Answer 1

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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.

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  • $\begingroup$ How exactly would this look, would I extend the single vector with W to another subspace? $\endgroup$
    – blub
    Nov 30, 2017 at 15:34
  • $\begingroup$ 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. $\endgroup$ Nov 30, 2017 at 15:38
  • $\begingroup$ How would I construct this space? $\endgroup$
    – blub
    Nov 30, 2017 at 22:07
  • $\begingroup$ 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. $\endgroup$ Dec 1, 2017 at 15:39

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