# 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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