# Proof of $\mathrm{dim}(U)\leq \mathrm{dim}(V)$ if $U$ is a subspace of $V$

I want to prove this classic theorem of linear algebra:

Let $V$ be a finite dimensional vector space, with $\mathrm{dim}(V)=n$. Let $U$ be a subspace of $V$ with $\mathrm{dim}(U)=m$, then $m\leq n$ and if $m=n$ then $U=V$.

I know there is the standard proof (e.g. Friedberg), who builds a maximal set of linearly independent vectors of $U$, then arguing that none linearly independent set of $V$ can contain more than $n$ elements, he concludes that this maximal set is a basis of $U$ and that it contains $m\leq n$ elements.

My alternative attempt: I was thinking about a simpler proof, so I thought in something like this.

Suppose $m\geq n$. Then a basis of $U$ has $m$ elements and a basis of $V$ has $n$ elements. Then there exists at least one vector $x \in U$ but $x \notin V$. Now, by the definition of subspace, $U$ must be a subset of $V$. Which is a contradiction due to the existence of at least this one element $x$. So $m\leq n$.

Question Is my attempt valid or at least the idea is correct? How can I improve it?

• First, you need to prove that $U$ has finite dimension. Suppose $m>$ and, by contradiction, check $m\leq n$. Why there exists at least one vector $x\in U$ but $x\notin V$? – Rafael Holanda Oct 21 '16 at 3:48
• Happens I've just written up a proof for my class: math.umn.edu/~dgrinber/4242/hw4.pdf (Proposition 0.1). – darij grinberg Oct 21 '16 at 4:02
• I am not sure how you get your contradiction. This result is trickier than it might seem, since it is not a-priori clear that $U$ is finite-dimensional. – darij grinberg Oct 21 '16 at 4:03
• @darijgrinberg: How are you defining finite-dimensional? You could take as definition that a space is finite-dimensional if it has a finite spanning set; now a suitable basis of $V$ must span $U,$ since $U$ is a subset of $V,$ and is finite, since $V$ is finite-dimensional. – Will R Oct 21 '16 at 4:36
• @WillR: Why would a suitable basis of $V$ span $U$ ? For it to span $U$, its entries have to lie in $U$ in the first place. – darij grinberg Oct 21 '16 at 6:13