Prove or give a counterexample: If $v_1, \dots, v_m$ is a linearly independent list of vectors in $V$ and $\lambda \in \mathbf F$ with $\lambda \ne 0$, then $\lambda v_1,\dots,\lambda v_n$ is linearly independent

Suppose

Since $\lambda \ne 0$ we can divide by lambda to get

Which implies $a_1 = a_2 = \dots = a_m = 0$. thus $(\lambda v_1, \dots, \lambda v_m)$ is linearly independent.