Problem. Let \(V\) be a finitely generated vector space such that \(\dim\ V=n\), and let \(\left(v_{1},\dots,v_{n}\right)\) be a sequence of vectors from \(V\). Prove that \(\left(v_{1},\dots,v_{n}\right)\) is linearly independent if and only if \(\left(v_{1},\dots,v_{n}\right)\) spans \(V\).
Proof. We prove the two implications separately.
\(\left(\Rightarrow\right)\):
Let us assume that \(\left(v_{1},\dots,v_{n}\right)\) is linearly independent. That means that the only solution to the equation \(a_{1}v_{1}+\dots+a_{n}v_{n}=0\) s.t \(a_{1},\dots,a_{n}\in\mathbb{F}\) is \(a_{1}=\dots=a_{n}=0\).
Let us assume for the sake of contradiction that \(\left(v_{1},\dots,v_{n}\right)\) doesn’t span \(V\).
In that case, \(\exists w\in V\ s.t\ w\notin\text{Span }\left(v_{1},\dots,v_{n}\right)\).
In other words, w is not a linear combination of this sequence.
Recall the following: if a vector is not a linear combination of a sequence, and if the sequence is linearly independent, then we can add it to the sequence and the new sequence will also be linearly independent - which means \(\left(v_{1},\dots,v_{n},w\right)\) is still linearly independent.
On the other hand, recall that if \(\dim\ V=n\) and a sequence of vectors is of length \(k>n\), then it must be linearly dependent.
We concluded that \(\left(v_{1},\dots,v_{n},w\right)\), a sequence of vectors from \(V\) of length \(n+1>n\) is still linearly independent, which is a contradiction; so \(\left(v_{1},\dots,v_{n}\right)\) must span \(V\), as required.
\(\left(\Leftarrow\right)\):
Let us assume that \(\left(v_{1},\dots,v_{n}\right)\) spans \(V\) and assume for the sake of contradiction that \(\left(v_{1},\dots,v_{n}\right)\) is linearly dependent.
Recall that if a sequence is linearly dependent, one of its vectors can be presented as a linear combination of the other vectors.
Then \(\exists1\le i\le n\) s.t \(v_{i}\) is a linear combination of \(\hat{v}=\left(v_{1},\dots,v_{i-1},v_{i+1},\dots,v_{n}\right)\).
Recall that if a sequence of vectors spans \(V\) and is linearly dependent, we can remove a vector that is a linear combination of the other vectors and the sequence will maintain its spanning property. Therefore, \(\hat{v}\) spans \(V\).
However, \(\hat{v}\) is of length \(n-1\), and another standard result specifies that for every sequence \(\bar{v}\) which spans \(V\), \(\dim\ V\le len(\bar{v})\). This means \(n\le n-1\) which is a contradiction; therefore \(\left(v_{1},\dots,v_{n}\right)\) must be linearly independent, as required. ∎
My thanks to one ordered 100-tuple