Problem. Let \(V\) be a vector space and let \(v_{1},\dots,v_{n},u,w\in V\) be such that the sequence \((v_{1},v_{2},\dots,v_{n},u)\) is linearly dependent, while the sequence \((v_{1},\dots,v_{n},w)\) is linearly independent.
Is it necessarily the case that \(w\notin\operatorname{Span}(v_{1},v_{2},\dots,v_{n},u)\)?
Proof. The sequence \((v_{1},\dots,v_{n},w)\) is linearly independent, which means that the sequence \((v_{1},\dots,v_{n})\) is also linearly independent (recall that if a sequence is linearly dependent one of the vectors is a linear combination of the others, hence a sequence that is already linearly dependent will stay linearly dependent when we add a vector to it).
Recall that if you add a vector to a linearly independent sequence which does not belong to the span of this sequence, you receive a new sequence which is still linearly independent, hence since \((v_{1},v_{2},\dots,v_{n},u)\) is linearly dependent we get that \[u\in\operatorname{Span}(v_{1},v_{2},\dots,v_{n})\]
Recall that if a vector belongs to the span of a sequence, the span of this sequence is equal to the span of this sequence along with said vector, hence
\[\operatorname{Span}(v_{1},v_{2},\dots,v_{n})=\operatorname{Span}(v_{1},v_{2},\dots,v_{n},u)\] Which means \(w\) cannot belong to \(\operatorname{Span}(v_{1},v_{2},\dots,v_{n},u)\), because in that \[w\in\operatorname{Span}(v_{1},v_{2},\dots,v_{n})\]
which would mean the sequence \((v_{1},\dots,v_{n},w)\) is linearly dependent. ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.