Prop: Let $v_1,...,v_n$ be a basis for the vector space $V$. And $T : V \rightarrow W$ is an isomorphism, then $T(v_1),..,T(v_n)$ is a basis for the vector space $W$.
To prove this proposition, I used the fact that if $v_1,...,v_n$ is a basis, then they must be linearly independent. So, $\sum \alpha_i v_i=0.$ Let $T(v_i) = w_i$, by definition of isomorphism, $T^{-1} : W \rightarrow V$ so $T^{-1}(w_i) = v_i$.
$$T(\sum \alpha_i v_i) = \sum \alpha_i T(v_i) = \sum \alpha _ i w_i \implies T(0) = \sum \alpha _ i w_i \implies \sum \alpha _ i w_i = 0 $$
Since we know this fact, we can now conclude that
$$\sum \alpha_i T(v_i) = \sum \alpha_i w_i \implies \sum \alpha_i T(v_i) = 0$$
And if $\sum \alpha_i T(v_i)=0$, then it is linearly independent so is a basis for $W \square$
Say we have $$\sum \alpha_i T(v_i) = 0$$ then we have$$\sum T(\alpha_i v_i) = 0\implies T(\sum \alpha_i v_i) = 0$$
Since $T$ is injective we have now $$\sum \alpha_i v_i = 0$$
and since $v_i$ are independent we have $\alpha _i=0$ for all $i$.