# Proving a vector space over itself have no subspaces

The question is:

Let $F$ be a field. $F$ is a vector space over itself which we'll represent as $V$. Find all the vector sub-spaces of $V$ and prove your answer.

After thinking about it, I'm pretty sure the only subspace of $V$ is $V$ itself. Since for every element we remove (Let's say $V_2 +V_3 = V_1$) the elements $V_2$ or $V_3$ can't be in the subspace either otherwise the closure axiom wouldn't hold, and the cycle continues.

But, after thinking a little longer, I have no idea how to prove it. I don't know how I can say that every element is the result of addition of two other elements. Also I have no idea how to prove this will result in no elements in the subspace.

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What are ideals of a field? – Alexei Averchenko Nov 19 '12 at 16:35
If $V=F$ is a field considered as a vector space over itself, $V$ is not the only subspace. You also have the other trivial subspace $(0)$. – Andrea Mori Nov 19 '12 at 17:05
Thanks for the edits! English is not my mother language so translating is sometimes difficult for me. – Nescio Nov 19 '12 at 17:37

Hint: Try and find a basis of $F$, which will give you it's dimension. The dimension of any subspace is less than or equal to the dimension of $F$. (And indeed, for any finite dimensional vector space there exists a (non-unique) subspace for any dimension smaller than the dimension of the original space.)

If $F$ has no subspaces other than itself (and $\{0\}$, which is a subspace of any vector space!) this tells us what the dimension is, and conversely, having this particular dimension tells us what the possible subspaces are!

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Hint: Vector subspaces are closed under multiplication by scalars. If $v$ is a non-zero element of $V$, can you produce any other element of $V$ through multiplication by a scalar?

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Ok, that did help me. I now know that if $1_F\in U$ it can't be a vector subspace. But I'm not sure what happens if $1_F\notin U$, since the axioms for vector spaces don't require a vector identity element, right? – Nescio Nov 19 '12 at 17:36
@Nescio: The hint wasn't to assume that $1_F\in U$, but that some $0\ne v\in U$. Is $v^{-1}v$ in $U$? – joriki Nov 19 '12 at 17:54
@Nescio: By the way, I think you mean that if $1_F\in U$, then $U$ can't be a proper vector subspace. $V$ itself is a vector subspace of $V$, but not a proper vector subspace. – joriki Nov 19 '12 at 17:57
In the end, after actually understanding how to use the dimensions, I found the other way simpler. But thanks for the help! – Nescio Nov 21 '12 at 11:29
@Nescio: A good example showing that the number of upvotes doesn't necessarily reflect the utility of the answer to the person who asked the question... – joriki Nov 21 '12 at 12:17