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 \(V\).