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\begin{document}
\title{{\Large Semester 2025B, Exam B, Q1\vspace{-2em}}}
\maketitle
\begin{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
$V$.
\end{problem*}
\begin{proof}
We prove the two implications separately.

\medskip{}

\textbf{$\left(\Rightarrow\right)$: }

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.

\medskip{}

\textbf{$\left(\Leftarrow\right)$: }

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.
\end{proof}
\begin{quote}
My thanks to one ordered 100-tuple 
\end{quote}

\end{document}
