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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.

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

\begin{align*}T(v_j) &= T(0v_1 + \dots + 0v_{j-1} + 1v_j + 0v_{j+1} + \dots + 0v_n)\\ &= 0w_1 + \dots + 0w_{j-1} + 1w_j + 0w_{j+1} + \dots + 0w_n\\ &= w_j.\end{align*}

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