Prove or give a counterexample: If $v_1,\dots,v_m$ is linearly independent in $V$, then

Is linearly independent

The (dumb) direct proof goes very similar to 2a/6

Suppose

$$a_1(5v_1-4v_2) + a_2v_2 + \dots + a_mv_m = 0$$

Then

$$5a_1 v_1 + (a_2 - 4a_1) v_2 + a_3v_3 + \dots + a_mv_m = 0$$

Implying $5a_1 = (a_2-4a_1)=a_3=\dots=a_m = 0$ by the independence of $v_1\dots,v_m$.
And $a_2=0$ since $(a_2-4a_1)+4a_1=a_2=0$.

This proof works, but has the minor problem of being utterly braindead.

The morally correct proof is noticing the given vectors span $V$ (same reason as in 2a/1) then applying 2.42, but sadly 2.42 hasn't been introduced yet so this would not be valid.