# Is a linear map $f: V \to W$ defined by images of the basis of $V$?

When I define a linear map $f: V \to W$, and $\{v_1, ..., v_n\}$ is $V$'s basis, is it sufficient to provide just $f(v_1), ..., f(v_n)$ because by Rank-Nullity, $f(v_1), ..., f(v_n)$ are distinct and basis of $W$?

At the same time I not sure whether I am applying Rank-Nullity in the right way, since Rank-Nullity assumes that we build basis of $V$ upon $N(f)$.

for any $v \in V$, we have (since $\{v_1, \dots, v_n\}$ is a base) $$f(v) = f(\alpha_1 v_1 + \dots + \alpha_n v_n) = \alpha_1 f(v_1) + \dots + \alpha_n f(v_n)$$
so if you specify $f(v_1), \dots, f(v_n)$ you are completely defining the linear map $f(v)$
The $f(v_i)$ need not be a basis or even distinct. At any rate $f$ is uniquely determined by its values on a basis (or even on a generating system): If $v\in V$ is arbitrary, we find $c_i$ such that $v=c_1v_1+\ldots +c_nv_n$ and obtain by linearity of $f$ that necessarily $$f(v)=f(c_1v_1+\ldots +c_nv_n)=c_1f(v_1)+\ldots c_nf(v_n).$$