设v1, v2, v3, v4张成V，证明v1 – v2, v2 – v3, v3 – v4, v4也张成V.

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 $$V$$.