Problem
$v_1, …, v_n$ is a basis of a vector space $V$ over $\mathbb{F}$.
Proof that $b_1v_1, …, b_nv_n$ for $b_1, …, b_n \in \mathbb{F} \backslash \{0\}$ is a basis of $V$.
Solution
The example solution is surprisingly long, proving technicalities that are (I thought) already given because we are scaling basis vectors. Since $b_i \ne 0$, none of the basis vectors $v_i$ can be dropped. Why is it not enough for a proof to just state that
$v_1, …, v_n$ is a basis for $V \Rightarrow$
$\langle v_1, …, v_n\rangle = V \land v_1, …, v_n$ are linearly independent $\Rightarrow$
$\forall i, 1 \le i \le n: b_i v_i \in \langle v_1, …, v_n \rangle \land b_i v_i \notin \langle v_1, …, v_{i-1}, v_{i+1}, …, v_n \rangle \Rightarrow$
$\langle b_1v_1, …, b_nv_n\rangle = V \land b_1v_1, …, b_nv_n$ are linearly independent $\Rightarrow$
$b_1v_1, …, b_nv_n $ is a basis for $V \ \Box$
Best Answer
Suppose that the the set of vectors $\{b_1v_1,...,b_nv_n\}$ is not linearly independent. Then there is $k\in\{1,...,n\}$ such that $v_kb_k$ is a linear combination of the other elements of the set, which means there are constants $\alpha_i$ not all zero such that $$v_kb_k=\sum_{i\neq k}\alpha_i(b_iv_i),$$ where the sum is over $\{1,...,n\}-\{k\}$. Since $b_k\neq0$, you can multiply this equation by $b_k^{-1}$ and use the fact that $\frac{\alpha_ib_i}{b_k}$ is a scalar in $F$ to deduce $v_k$ is can be written as linear combination of the other $v_i$. This would contradict the fact that $\{v_1,...,v_n\}$ is a basis of $V$, so it must be false. Thus these vectors are linearly independent. Finally, any element $u$ of $V$ can be written in the form $$u=\sum_{i=1}^na_iv_i=\sum_{i=1}^nc_i(b_iv_i),$$ where $c_i=a_i/b_i$. Since the $b_i$ are assumes to be zero, $c_i$ always exists. So this set does indeed span $V$ and must be a basis. As others pointed out in the comments, you only proved the first part (the linear independence.)