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  "body": "\n## Proof\n\nWe compute the image of each basis vector of $U$ under $T \\circ S$ in two steps.\n\n1. Write $A = (a_1, \\dots, a_p)$, $B = (b_1, \\dots, b_n)$ and $C = (c_1, \\dots, c_m)$ for the [[bases|0KCQ9Y3QKA]] of $U$, $V$ and $W$.\n2. Let $[S]_{B,A} = (s_{jk})$, an $n \\times p$ matrix. By the definition of the [[matrix of a linear map|0YQSGNYE3K]], $S(a_k) = \\sum_{j=1}^n s_{jk} b_j$ for $1 \\le k \\le p$.\n3. Let $[T]_{C,B} = (t_{ij})$, an $m \\times n$ matrix. Then $T(b_j) = \\sum_{i=1}^m t_{ij} c_i$ for $1 \\le j \\le n$.\n4. The [[composition|00BKRZKMSS]] $T \\circ S$ is linear: $T(S(u + \\lambda u')) = T(S(u)) + \\lambda T(S(u'))$ for $u, u' \\in U$ and $\\lambda \\in F$, since $S$ and $T$ are [[linear|0NRYHJPQG2]].\n5. Fix $k$ with $1 \\le k \\le p$. By step 2 and the linearity of $T$, $(T \\circ S)(a_k) = T\\left(\\sum_{j=1}^n s_{jk} b_j\\right) = \\sum_{j=1}^n s_{jk} T(b_j)$.\n6. By step 3, this equals $\\sum_{j=1}^n s_{jk} \\sum_{i=1}^m t_{ij} c_i = \\sum_{i=1}^m \\left(\\sum_{j=1}^n t_{ij} s_{jk}\\right) c_i$.\n7. By the definition of the [[matrix product|032Q9FB8MT]], $\\sum_{j=1}^n t_{ij} s_{jk}$ is the entry $(i, k)$ of $[T]_{C,B}\\,[S]_{B,A}$.\n8. The coordinates of a vector in the basis $C$ are unique. So, by step 6, entry $(i, k)$ of $[T \\circ S]_{C,A}$ is $\\sum_{j=1}^n t_{ij} s_{jk}$.\n9. By steps 7 and 8, the two $m \\times p$ matrices have the same entries. Hence $[T \\circ S]_{C,A} = [T]_{C,B}\\,[S]_{B,A}$.\n\n## Idea\n\nColumn $k$ of a matrix records where the $k$-th basis vector goes. Pushing $a_k$ first through $S$ and then through $T$ produces exactly the sum that defines the $(i, k)$ entry of the product. This is why matrix multiplication is defined the way it is.\n",
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