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\begin{document}
\title{{\Large Semester 2026A, Exam A, Q4\vspace{-2em}}}

\maketitle
\textbf{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 $w\notin\operatorname{Span}(v_{1},v_{2},\dots,v_{n},u)$?

\medskip{}

\rule[0.5ex]{1\columnwidth}{1pt}

\medskip{}

\begin{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.
\end{proof}
\begin{quote}
This is an exam problem I translated and solved. The original exam
was written by the Linear Algebra 1 course lecturers at HUJI.
\end{quote}

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