【归档】设v1, v2, v3, v4张成V，证明v1 - v2, v2 - v3, v3 - v4, v4也张成V.

Note: 旧的wordpress博客弃用，于是将以前的笔记搬运回来。

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Question:
Suppose $v_1,v_2,v_3,v_4$ spans $V$.
Prove that $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, $v_4$ also spans $V$.
Solution:
Let $v\in V$. Since $(v_1,v_2,v_3,v_4)$ spans $V$, we know that there exist scalars $a_1,a_2,a_3,a_4\in \mathbb{F}$ such that:
$a_1{v}_1+a_2{v}_2+a_3{v}_3+a_4{v}_4=v$.
We seek coefficient $b_1,b_2,b_3,b_4\in \mathbb{F}$ such that:
$b_1(v_1-v_2)+b_2(v_2-v_3)+b_3(v_3-v_4)+b_4{v}_4=v=a_1{v}_1+a_2{v}_2+a_3{v}_3+a_4{v}_4$.
Using distributive property, we get:
$b_1{v}_1+(b_2-b_1)v_2+(b_3-b_2)v_3+(b_4-b_3)v_4=a_1{v}_1+a_2{v}_2+a_3{v}_3+a_4{v}_4$.
So we get:
$b_1=a_1,b_i-b_{i-1}=a_i$. $(i\in [2,4])$
Now we claim that for each $b_i$, there exist $b_i={\sum }_{k=1}^i{a}_k$. Clearly this is correct for $b_1$, since $b_1=a_1$. Now we assume $j\in [2,3]$ and $b_j={\sum }_{k=1}^j{a}_k$. Since $b_i-b_{i-1}=a_i$, we can deduce:
$b_{j+1}=b_j+a_{j+1}={\sum }_{k=1}^j{a}_k+a_{j+1}={\sum }_{k=1}^{j+1}{a}_k$.
So the formula is proved by induction.
This proved that:
$a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)(v_4-v_3)+(a_1+a_2+a_3+a_4)v_4=v$.
Since for each $v\in span(v_1,v_2,v_3,v_4)$ there exist this formula, the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ spans $V$.