Suppose $v_1, v_2, v_3, v_4$ is linearly independent in $V$. Prove that the list

Is also linearly independent.

This is the same list as in 2a/1, notice the two lists are "basically the same" (meaning we can get from one to the other and vise versa by taking linear combinations), so their equivilance makes sense.

The proof is quite straightforward, suppose

Then distribute to get

Since $(v_1,\dots,v_4)$ is linearly independent, this implies

Which, using the same telescoping trick as before implies $a_2 = a_3 = a_4 = 0$ (add $a_1$ to $(a_2-a_1)$, etc). This completes the proof.

Looking back I don't like this proof, it's a direct computation which doesn't generalize the intuitive concept I stated at the top.

Let's state the general theorem: If $v_1,\dots,v_n$ are linearly independent in $V$ and $w_1,\dots,w_n$ are such that for all $1 \le k \le n$ there is some choice of $a_1,\dots,a_n$ so that

Then $(w_1,\dots,w_n)$ is linearly independent in $V$.

Proving this is easy once we have access to the tools in 2.B, specifically it's a direct consequence of 2.42.