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  "body": "\n## Proof\n\n1. Write $B = (b_1, \\dots, b_n)$ and $C = (c_1, \\dots, c_m)$. Let $B^* = (b_1^*, \\dots, b_n^*)$ and $C^* = (c_1^*, \\dots, c_m^*)$ be the [[dual bases|0W6M3PSD08]], so $b_l^*(b_j) = \\delta_{lj}$ and $c_i^*(c_k) = \\delta_{ik}$.\n2. Here $\\delta_{ij} = 1$ if $i = j$ and $\\delta_{ij} = 0$ otherwise.\n3. Let $[T]_{C,B} = (a_{ij})$, an $m \\times n$ matrix. By the [[definition of the matrix of a linear map|0YYMY95FWS]], $T(b_j) = \\sum_{k=1}^{m} a_{kj} c_k$ for $1 \\le j \\le n$.\n4. The [[dual map|0C2TQMW5A1]] is $T^* \\colon W^* \\to V^*$, $T^*(f) = f \\circ T$. Its matrix $[T^*]_{B^*, C^*} = (d_{li})$ is $n \\times m$.\n5. By the same definition, $T^*(c_i^*) = \\sum_{l=1}^{n} d_{li} b_l^*$ for $1 \\le i \\le m$.\n6. Fix $i \\le m$ and $j \\le n$. Evaluating step 5 at $b_j$ gives $T^*(c_i^*)(b_j) = \\sum_{l=1}^{n} d_{li} \\delta_{lj} = d_{ji}$.\n7. On the other hand, by step 4 and step 3, $T^*(c_i^*)(b_j) = c_i^*(T(b_j)) = \\sum_{k=1}^{m} a_{kj} c_i^*(c_k)$.\n8. Since $c_i^*(c_k) = \\delta_{ik}$, the sum in step 7 equals $a_{ij}$.\n9. By steps 6 to 8, $d_{ji} = a_{ij}$ for all $i \\le m$ and $j \\le n$.\n10. By the definition of the [[transpose|0W809K6AW0]], step 9 says $[T^*]_{B^*, C^*} = \\left([T]_{C,B}\\right)^T$.\n\n## Idea\n\nThe entries of both matrices can be read off by pairing a dual basis form with a basis vector. The number $c_i^*(T(b_j))$ is entry $(i, j)$ of $[T]$ and also entry $(j, i)$ of $[T^*]$, so the two matrices are transposes.\n",
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