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  "body": "\n## Proof\n\nWe first prove uniqueness, then build the map from the coordinates of each vector.\n\n1. Since $(b_i)_{i \\in I}$ is a [[basis|0KCQ9Y3QKA]], each $v \\in V$ is a [[linear combination|0V8F3E48E6]] $v = \\sum_{i \\in I} a_i b_i$ in exactly one way. Here $a_i \\in F$, and $a_i = 0$ for all but finitely many $i$.\n2. Let $S, T \\colon V \\to W$ be [[linear maps|0NRYHJPQG2]] with $S(b_i) = T(b_i)$ for all $i \\in I$. Let $v \\in V$ be written as in step 1.\n3. The sum in step 1 has finitely many nonzero terms. Linearity, applied once for each term, gives $S(v) = \\sum_{i \\in I} a_i S(b_i)$ and $T(v) = \\sum_{i \\in I} a_i T(b_i)$.\n4. The right sides of step 3 are equal term by term. So $S(v) = T(v)$ for all $v \\in V$, and $S = T$.\n5. Now let $(w_i)_{i \\in I}$ be a family of vectors of $W$. For $v = \\sum_{i \\in I} a_i b_i$ as in step 1, define $T(v) = \\sum_{i \\in I} a_i w_i$.\n6. This is a finite sum, since only finitely many $a_i$ are nonzero. The coefficients $a_i$ are unique, so $T(v)$ is well defined.\n7. $T$ is additive. Let $v = \\sum_i a_i b_i$ and $v' = \\sum_i a'_i b_i$. Then $v + v' = \\sum_i (a_i + a'_i) b_i$, and by uniqueness these are the coefficients of $v + v'$.\n8. Hence $T(v + v') = \\sum_i (a_i + a'_i) w_i = T(v) + T(v')$.\n9. $T$ is homogeneous. For $c \\in F$, $cv = \\sum_i (c a_i) b_i$, and by uniqueness these are the coefficients of $cv$. Hence $T(cv) = \\sum_i c a_i w_i = c\\, T(v)$.\n10. So $T$ is linear. For $j \\in I$, the vector $b_j$ has coefficient $1$ at $j$ and $0$ elsewhere, by uniqueness. Hence $T(b_j) = w_j$.\n11. If $T'$ is another linear map with $T'(b_j) = w_j$ for all $j \\in I$, then $T'$ and $T$ agree on the basis. By step 4, $T' = T$.\n12. So there is exactly one linear map $T \\colon V \\to W$ with $T(b_i) = w_i$ for all $i \\in I$.\n\n## Idea\n\nEvery vector is a unique combination of basis vectors, and a linear map must respect combinations. So the images of the basis vectors fix the map. Conversely, any choice of images defines a map by sending each combination to the same combination of the chosen images.\n",
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