Steinitz exchange lemma

Steinitz exchange lemma If $\{v_1,\dots ,v_m\}$ is a set of linearly independent vectors in vector space V, and $\{w_1,\dots ,w_n\}$ spans V, then:

1. Possibly after reordering the $w_i$, $\{v_1,\dots ,v_m,w_{m+1},\dots ,w_n\}$ spans V.
2. $m\le n$

1. $H_m$: Steinitz exchange lemma holds for all non-negative integers m.
2. $H_0$: Steinitz exchange lemma holds when $m=0$ $\mathrm{\varnothing }$ is a set of linearly independent vectors in vector space V. Based on the conditions given $\{w_1,\dots ,w_n\}$ spans V. And clearly, $m=0\le n$
3. $H_k$: Steinitz exchange lemma holds for some non-negative integers $k$
4. $H_{k+1}$: Steinitz exchange lemma holds for $(k+1)$. Based on $H_k$, $\{v_1,\dots ,v_k,w_{k+1},\dots ,w_n\}$ spans V.
   $$v_{k+1}\in \mathbf{\text{V}}\iff v_{k+1}=\sum _{i=1}^k{\lambda }_i{v}_i+\sum _{i=k+1}^n{\lambda }_i{w}_i$$

   Since $v_1,\dots ,v_{k+1}$ are linearly independent,

   $$\begin{array}{rl}\Rightarrow & v_{k+1}-\sum _{i=1}^k{\lambda }_i{v}_i\ne 0\\ \Rightarrow & v_{k+1}\ne \sum _{i=1}^k{\lambda }_i{v}_i\\ \Rightarrow & \text{At least one of }{\lambda }_{k+1},\dots {\lambda }_n\text{is non-zero}\end{array}$$

   Rearrange $w_i$ so that ${\lambda }_{k+1}\ne 0$

   $$\begin{array}{rl}\Rightarrow & v_{k+1}=\sum _{i=1}^k{\lambda }_i{v}_i+{\lambda }_{k+1}{w}_{k+1}+\sum _{i=k+2}^n{\lambda }_i{w}_i\\ \iff & w_{k+1}=\frac{1}{-{\lambda }_{k+1}}(\sum _{i=1}^k{\lambda }_i{v}_i-v_{k+1}+\sum _{i=k+2}^n{\lambda }_i{w}_i)\end{array}$$

   $\mathrm{\forall }v\in$V, $\mathrm{\exists }$ a linear combination of$\{v_1,\dots ,v_k,w_{k+1},\dots ,w_n\}=v$ Now we substitute $w_{k+1}=\frac{1}{-{\lambda }_{k+1}}(\sum _{i=1}^k{\lambda }_i{v}_i-v_{k+1}+\sum _{i=k+2}^n{\lambda }_i{w}_i)$ into the linear combination. $\Rightarrow \mathrm{\forall }v\in$V, $\mathrm{\exists }$ a linear combination of$\{v_1,\dots ,v_k,v_{k+1},w_{k+2},\dots ,w_n\}=v$ Therefore $H_{k+1}$ holds, if $H_k$ is true.

5. Based on M.I., $H_m$ is true.