Vector space/Basis/Exchange lemma/Fact/Proof

We show first that the new family is a generating system. Because of

${\displaystyle {}w=\sum _{i=1}^{n}s_{i}v_{i}\,}$

and ${\displaystyle {}s_{k}\neq 0}$, we can express the vector ${\displaystyle {}v_{k}}$ as

${\displaystyle {}v_{k}={\frac {1}{s_{k}}}w-\sum _{i=1}^{k-1}{\frac {s_{i}}{s_{k}}}v_{i}-\sum _{i=k+1}^{n}{\frac {s_{i}}{s_{k}}}v_{i}\,.}$

Let ${\displaystyle {}u\in V}$ be given. Then, we can write

${\displaystyle {}{\begin{aligned}u&=\sum _{i=1}^{n}t_{i}v_{i}\\&=\sum _{i=1}^{k-1}t_{i}v_{i}+t_{k}v_{k}+\sum _{i=k+1}^{n}t_{i}v_{i}\\&=\sum _{i=1}^{k-1}t_{i}v_{i}+t_{k}{\left({\frac {1}{s_{k}}}w-\sum _{i=1}^{k-1}{\frac {s_{i}}{s_{k}}}v_{i}-\sum _{i=k+1}^{n}{\frac {s_{i}}{s_{k}}}v_{i}\right)}+\sum _{i=k+1}^{n}t_{i}v_{i}\\&=\sum _{i=1}^{k-1}{\left(t_{i}-t_{k}{\frac {s_{i}}{s_{k}}}\right)}v_{i}+{\frac {t_{k}}{s_{k}}}w+\sum _{i=k+1}^{n}{\left(t_{i}-t_{k}{\frac {s_{i}}{s_{k}}}\right)}v_{i}.\end{aligned}}}$

To show the linear independence, we may assume ${\displaystyle {}k=1}$ to simplify the notation. Let

${\displaystyle {}t_{1}w+\sum _{i=2}^{n}t_{i}v_{i}=0\,}$

be a representation of ${\displaystyle {}0}$. Then

${\displaystyle {}0=t_{1}w+\sum _{i=2}^{n}t_{i}v_{i}=t_{1}{\left(\sum _{i=1}^{n}s_{i}v_{i}\right)}+\sum _{i=2}^{n}t_{i}v_{i}=t_{1}s_{1}v_{1}+\sum _{i=2}^{n}{\left(t_{1}s_{i}+t_{i}\right)}v_{i}\,.}$

From the linear independence of the original family, we deduce ${\displaystyle {}t_{1}s_{1}=0}$. Because of ${\displaystyle {}s_{1}\neq 0}$, we get ${\displaystyle {}t_{1}=0}$. Therefore, ${\displaystyle {}\sum _{i=2}^{n}t_{i}v_{i}=0}$, and hence ${\displaystyle {}t_{i}=0}$ for all ${\displaystyle {}i}$.