Answer
It can be proven by using Theorem 4.12
Work Step by Step
Suppose that $S=\left\{{v}_{1}, {v}_{2}, \cdots, {v}_{n}\right\}$ is a basis for $V$ and $c$ is a nonzero scalar. Since $S$ is a basis for $V$, $ V$ has dimension $n $. Since $S_{1}$ has $n$ vectors, then by Theorem 4.1, we have to prove that $S_1$ is independent set of vectors. So assume that $a_{1}, \cdots, a_{n}$ are
scalars such that $a_{1}\left(c v_{1}\right)+a_{2}\left(c v_{2}\right)+\cdots+a_{n}\left(c v_{n}\right)=0$, then we get
$$
\left(c a_{1}\right) v_{1}+\left(c a_{2}\right) v_{2}+\cdots+\left(c a_{n}\right) v_{n}=0
$$
Since $S$ is independent, we must have
$$
c a_{1}=0, c a_{2}=0, \cdots, c a_{n}=0
$$
and since $c$ is not zero, these conditions imply that
$$
a_{1}=0, a_{2}=0, \cdots, a_{n}=0
$$
Therefore $S_{1}$ is independent, hence a basis.
