# Linear maps: map construction

From "Linear algebra done right" chapter 3:

Linear maps can be constructed that take on arbitrary values on a basis. Specifically, given a basis $\{v_1, \dots, v_n\}$ of $V$ and any choice of vectors $w_1, \dots, w_n \in W$, we can construct a linear map $T: V \to W$ such that $T(v_j) = w_j$ for $j = 1, \dots, n$. There is no choice of how to do this — we must define $T$ by $T(a_1v_1 + \dots + a_nv_n) = a_1w_1 + \dots + a_nw_n$, where $a_1, \dots, a_n$ are arbitrary elements of $\mathbb{F}$. Because $\{v_1, \dots, v_n\}$ is a basis of $V$, the equation above does indeed define a function $T$ from $V$ to $W$. You should verify that the function $T$ defined above is linear and that $T(v_j) = w_j$ for $j = 1, \dots, n$.

It seems obvious, and we can prove $\{w_1, \dots, w_j\}$ is a basis for $W$, but how to prove $T(v_j) = w_j$ exactly? Thank you.

• $v_j=0v_1+0v_2+\cdot+1v_j+0v_{j+1}+\cdots+0v_n$. Nov 15 '13 at 5:48
• Got it. Thanks. Nov 16 '13 at 14:57
• Good. You can write it up as an answer, help clear up the Unanswered Questions list. Nov 17 '13 at 4:25

First of all, you can't prove that $\{w_1, \dots, w_n\}$ is a basis for $W$. This is false as $\{w_1, \dots, w_n\}$ is any choice of $n$ vectors in $W$, they may not even be distinct. For example, the construction outlined above works perfectly well for $w_1 = \dots = w_n = 0$.
Regarding your question, as Gerry Myerson pointed out in his comment, given $T : V \to W$ defined by
$$T(a_1v_1 + \dots + a_nv_n) = a_1w_1 + \dots + a_nw_n$$
we can see that $T(v_j) = w_j$ by noting that