# Prove any linearly indpendent set with n vectors is a basis for the space

There is the following lemma:

Any linearly independent set can be expanded to make a basis

Also it's well-known that all the space's basis has the same amount of vectors ($n$).

But how can be proved that any linearly independent set in the space with the length $n$ is able to span that space (i.e. is a basis)?

The question is about finite-dimensional vector space.

• You might want to restrict your claims to finite-dimensional vector spaces, or else some of these things get rather complex (like "what's the right definition of 'basis'?") – John Hughes Apr 1 '17 at 15:20
• @John Hughes I edited my question – Mergasov Apr 1 '17 at 15:23

Suppose not. Call the set of $n$ independent vectors $S$. If $S$ spans the vector space ($V$), then it's a basis, But since we're assuming not, there's some vector $v$ not in $span(S)$.
In that case, $S \cup \{v\}$ is linearly independent. (This requires proof, which you should provide: assume that it's not linearly independent, conclude something about the coefficient of $v$, and then work from there...)
But that means that the dimension of $V$ is at least $n+1$, for it has an $n+1$ element independent set.