Problem
Prove or give counterexample: If $v_1, v_2, \dots, v_m$ is a linearly independent list of vectors in $V$, then $$5v_1 - 4v_2, v_2, v_3, \dots, v_m$$ is linearly independent.
Solution
By the linear independence of $v_1, v_2, \dots, v_m$, we must have that $$0 = a_1v_1 + a_2v_2 + \dots + a_mv_m$$ $$=\frac{a_1}{5}\left(5v_1 - 4v_2\right) + \left(a_1\frac{9}{4} + a_2\right)v_2 + a_3 v_3 + \dots + a_m v_m$$ implies $a_1 = a_2 = \dots = a_m = 0$ which in turn implies $\frac{a_1}{5} = a_1\frac{9}{4} + a_2 = a_3 = \dots = a_m = 0$, thus the list $5v_1 - 4v_2, v_2, v_3, \dots, v_m$ is linearly independent if the list $v_1, v_2, v_3, \dots, v_m$ is.