# On a proof of dimension of vector spaces

There is a proof in my notes showing that if $U$ is a subspace of a finite-dimensional vector space $V$ then $\dim U \le \dim V$. It proceeds by extending a basis of $U$ to a basis of $V$. Wouldn't it be shorter and neater to say if $\dim U > \dim V$ then there would be at least one vector $u$ not in the span of $v_1, \ldots, v_n$. But then $u$ would lie outside $V$ which is impossible?

Or is it not correct to argue like this?

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It is pretty much the same thing in both cases. – Mlazhinka Shung Gronzalez LeWy Jun 26 '13 at 20:39

I'll limit the discussion to finite dimensional spaces, which seems what the question deals with.

The main point in the theorem is proving that also $U$ is finitely generated, which isn't obvious and in fact doesn't hold in more general situations like modules over non noetherian rings.

Without knowing the proof in your notes it's difficult to guess the order of arguments. Here's how I usually present it.

1. The exchange lemma guarantees that two bases of $V$ have the same number of elements (which is defined to be $\dim V$) and any linearly independent set in $V$ has at most $\dim V$ elements.

2. By induction, if no finite set of elements of $U$ spans $U$, we can build a linearly independent set in $U$ having $n$ elements, for all $n$.

• The fact is clear for $n=0$, because the empty set is linearly independent.

• If $\{u_1,\dots,u_n\}$ is linearly independent, since by hypothesis it doesn't span $U$, there exists $u_{n+1}\in U\setminus\operatorname{span}\{u_1,\dots,u_n\}$ and it's easy to see that $\{u_1,\dots,u_n,u_{n+1}\}$ is linearly independent.

3. Therefore the hypothesis that $U$ is not finitely generated leads to a contradiction.

4. From any finite set of generators of $U$ we can extract a basis which, by the exchange lemma mentioned above, has at most $\dim V$ elements. Thus $U$ is finite dimensional and $\dim U\le\dim V$.

Your argument seems to be as follows: $U$ cannot contain an infinite linearly independent set; therefore a maximal linearly independent set in $U$ is finite. Now, the number of its elements cannot exceed $\dim V$, because of the exchange lemma. A maximal linearly independent set spans $U$, otherwise (with the same argument as in the induction step above) it could be extended. In this form the argument is correct.

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Conceptually, these things are different.

Saying that $U$ is a subset of $V$ and then building up is quite natural.

Starting from the point that $\dim U > \dim V$ does not tell you anything. There is no connection between $U$ and $V$. They can be arbitrary vector spaces. You would need to start talking about linear maps between the basis of $U$ and the basis of $V$.

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So the argument is wrong? – newb Jun 26 '13 at 20:45
But he (@newb) knows from the assumptions that $U\subseteq V$. – P.. Jun 26 '13 at 20:48
It's not that it's "wrong", but it's back-to-front. If $\dim U > \dim V$ then what is $U$? It can't be a subspace of $V$. So perhaps $V$ is a subspace of $U$? But the question says that $U$ is a subspace of $V$. I feel dizzy already. – Fly by Night Jun 26 '13 at 20:48

One advantage of the direct approach your text uses is that it can be used in a more general context. Working without the axiom of choice, but with Hall's Marriage Theorem (which is known to be implied by the ultrafilter principle), I believe this approach can prove that if $U$ is a (possibly infinite-dimensional) subspace of $V$, $L$ is a linearly independent set in $U$, and $S$ spans $V$, then $|L|\le|S|$, whereas your approach would require assuming that $|L|$ and $|S|$ are comparable.

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