Is the following proof correct?
Theorem. Given that $\alpha_1,\alpha_2,...,\alpha_n$ is a a basis for a vector space $V$ over a field $\mathbf F$, it then follows that the list of vectors $\lambda_1\alpha_1,\lambda_2 \alpha_2,...,\lambda_n \alpha_n$ is also a basis for $V$, where $\forall j\in I=\{1,2,...,n\}(\lambda_j\neq 0).$
Proof. Let $w$ be an arbitrary vector in $V$ and assume that for some $\beta_1,\beta_2,...,\beta_n\in\mathbf{F}$ it is the case that $$w = \sum_{j=1}^{n}\beta_j(\lambda_j\alpha_j)\tag{1}$$ now let $\eta_1,\eta_2,...,\eta_n$ be arbitrary scalars in $\mathbf{F}$ such that the following is true $$w = \sum_{j=1}^{n}\eta_j(\lambda_j\alpha_j)\tag{2}$$ subtracting $(1)$ from $(2)$ we have $$0 = \sum_{j=1}^{n}(\eta_j\lambda_j-\beta_j\lambda_j)\alpha_j\tag{3}$$ since the list of vectors $\alpha_1,\alpha_2,...,\alpha_n$ is a basis for $V$ it follows that given any $j\in I$ we have $$\eta_j\lambda_j-\beta_j\lambda_j =(\eta_j-\beta_j)\lambda_j= 0\tag{4}$$ since $\lambda_j\neq 0$ dividing $(4)$ by $\lambda_j$ implies $\beta_j=\eta_j$, thus $w$ can be represented uniquely as linear combination of the vectors $\lambda_1\alpha_1,\lambda_2\alpha_2,...,\lambda_n\alpha_n$ and since $w$ was arbitrary it follows that the above list is a basis for $V$.
$\blacksquare$
