Vector space/Basis/Exchange lemma/Fact/Proof

Proof

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

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

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

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

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

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

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

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

be a representation of ${}0$. Then

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

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

To fact