# Vector space and maximal proper subspace

• Feb 1st 2009, 03:02 PM
vincisonfire
Vector space and maximal proper subspace
Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.
Show that there is a subspace W of V maximal subject to the condition
that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .
(Do NOT assume that V is ﬁnite-dimensional.)
For the first part I thought that I could say :
1. Prove that we can exchange one of the vector in basis of V by $\displaystyle \vec{v}$.
2. Prove that the basis (B) is still one.
3.Let $\displaystyle W = B \backslash \vec{v}$.
4. $\displaystyle B \backslash \vec{v}$ is a basis and thus a maximal independant subset that doesn't contain $\displaystyle \vec{v}$.
5. W is a subspace of V.
Is it correct?
For the second part I don't have a good idea of what to do.
Any hint would be great. Thanks!
• Feb 1st 2009, 03:39 PM
NonCommAlg
Quote:

Originally Posted by vincisonfire
Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.
Show that there is a subspace W of V maximal subject to the condition
that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .
(Do NOT assume that V is ﬁnite-dimensional.)

the first part is a quick result of Zorn's lemma: let $\displaystyle C$ be the set of all subspaces W of V such that $\displaystyle v \notin W.$ since $\displaystyle v \neq 0,$ the set $\displaystyle C$ contains the 0 subspace and hence it's not empty.

choose any chain (totally ordered collection) of elements of $\displaystyle (C, \subseteq).$ then the union of those elements is still in $\displaystyle C.$ thus $\displaystyle C$ has a maximal element by Zorn's lemma.

for the second part, suppose $\displaystyle W$ is a maximal element of $\displaystyle C$ and $\displaystyle W \subsetneq W_1,$ for some subspace $\displaystyle W_1$ of $\displaystyle V.$ the claim is that $\displaystyle W_1=V$: by maximality of $\displaystyle W$ in $\displaystyle C,$ we must have $\displaystyle v \in W_1.$

thus $\displaystyle W+<v> \subseteq W_1.$ so we only need to show that $\displaystyle W+<v>=V.$ suppose $\displaystyle W+<v> \neq V,$ and let $\displaystyle v_1 \notin W + <v>.$ then $\displaystyle v \notin W + <v_1>$ and thus $\displaystyle W + <v_1> \in C,$ which

contradicts maximality of $\displaystyle W$ in $\displaystyle C. \ \Box$
• Feb 2nd 2009, 06:25 AM
vincisonfire
The version of Zorn's lemma I know says that $\displaystyle C$ needs to be closed under union of chains.
Here it is
Let U be any set, and B any nonempty collection of subsets of U. Suppose tnat B is closed under unions of chains, then B has a maximal element.

How is define the union of these chains?
• Feb 2nd 2009, 05:29 PM
NonCommAlg
Quote:

Originally Posted by vincisonfire

How is define the union of these chains?

if $\displaystyle \{W_{\alpha}: \ \alpha \in I \}$ is a chain of subspaces (under inclusion), then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha}$ is again a subspace. if $\displaystyle \forall \alpha \in I: \ W_{\alpha} \in C,$ as defined in my solution, then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha} \in C.$