I want to prove this classic theorem of linear algebra:

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

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

My alternative attempt:

I was thinking about a simpler proof, so I thought in something like this.

Suppose m≥nm\geq n. Then a basis of UU has mm elements and a basis of VV has nn elements. Then there exists at least one vector x∈Ux \in U but x∉Vx \notin V. Now, by the definition of subspace, UU must be a subset of VV. Which is a contradiction due to the existence of at least this one element xx. So m≤nm\leq n.

Question

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

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First, you need to prove that UU has finite dimension. Suppose m>m> and, by contradiction, check m≤nm\leq n. Why there exists at least one vector x∈Ux\in U but x∉Vx\notin V?

– Rafael Holanda

Oct 21 at 3:48

1

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 at 4:02

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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 UU is finite-dimensional.

– darij grinberg

2 days ago

1

@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 VV must span U,U, since UU is a subset of V,V, and is finite, since VV is finite-dimensional.

– Will R

2 days ago

1

@WillR: Why would a suitable basis of VV span UU ? For it to span UU, its entries have to lie in UU in the first place.

– darij grinberg

2 days ago

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