Prove the dimension of span of vectors v1,…,vmv_1,\dots, v_m in Rn\mathbb{R}^n is at most m

Prove dim(span(v1,…,vm))≤mv_1,\dots, v_m)) \leq m for v1,…,vm∈Rnv_1,\dots, v_m \in \mathbb{R}^n. I think one can prove it neatly as follows, but I am not sure if it is correct or rigorous. Let V = span(v1,…,vm), so the vectors v1,…,vm span V. By Steinitz exchange lemma, we can have at most mm linearly independent vectors v1,…,vmv_1, . . . , v_m in V , that is at most mm basis vectors. By the definition of dimension this
means that the dimension of V is at most mm so dim(span(v1,…,vm))≤mv_1,\dots, v_m)) \leq m.

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It’s correct !!
– Reinaldo R.
2 days ago

  

 

From any spanning set you can extract a basis.
– egreg
2 days ago

  

 

We n≤mn \leq m l.i vectors in VV, any another vector is a linear combination of this ones. So, dimV=n\dim V = n
– Reinaldo R.
2 days ago

  

 

Nice using Steinitz but non-necssary. You can also used an easy proof by contradiction. Try it as exercise, it is very easy.
– Piquito
2 days ago

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1 Answer
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You don’t need Steinitz, but using it is fine.

If one of the vectors in {v1,…,vm}\{v_1,\dots,v_m\} is a linear combination of the others (that is, the set is linearly dependent), removing it from the set doesn’t change the span (prove it).

Continue until you have to stop because the set is linearly independent. Then you have a basis, which obviously has at most mm elements.