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  "body": "\n## Proof\n\nWe compare column spaces, and then pass to transposes for the second inequality.\n\n1. Let $A = (a_{ij})$ be $m \\times n$ and $B = (b_{jk})$ be $n \\times p$. Let $\\alpha_1, \\dots, \\alpha_n \\in F^m$ be the columns of $A$.\n2. By the definition of the [[matrix product|032Q9FB8MT]], entry $i$ of column $k$ of $AB$ is $\\sum_{j=1}^n a_{ij} b_{jk}$. So column $k$ of $AB$ equals $\\sum_{j=1}^n b_{jk} \\alpha_j$.\n3. Hence each column of $AB$ lies in the [[column space|03AJKXH1AV]] $\\operatorname{Col}(A)$. Since $\\operatorname{Col}(A)$ is a subspace, $\\operatorname{Col}(AB) \\subseteq \\operatorname{Col}(A)$.\n4. $F^m$ has dimension $m$ by [[the dimension of the coordinate space|0QK7FPE6VE]], so $\\operatorname{Col}(A)$ is finite-dimensional by the [[dimension of a subspace|0QW084460E]].\n5. Applying the dimension of a subspace inside $\\operatorname{Col}(A)$ gives $\\dim \\operatorname{Col}(AB) \\le \\dim \\operatorname{Col}(A)$. By the definition of [[rank|0ZC9YTXZRY]], $\\rank(AB) \\le \\rank A$.\n6. Step 5 holds for any two matrices whose product is defined.\n7. By [[row rank equals column rank|0TZAG60ZYT]], $\\rank(AB) = \\rank\\left((AB)^T\\right)$, with the [[transpose|09DXWYRD43]].\n8. By the [[transpose of a product|09GK5PS824]], $(AB)^T = B^T A^T$, where $B^T$ is $p \\times n$ and $A^T$ is $n \\times m$.\n9. By step 6 applied to $B^T$ and $A^T$, $\\rank(B^T A^T) \\le \\rank B^T$.\n10. By row rank equals column rank again, $\\rank B^T = \\rank B$. So $\\rank(AB) \\le \\rank B$.\n11. By steps 5 and 10, $\\rank(AB) \\le \\min(\\rank A, \\rank B)$.\n\n## Idea\n\nEvery column of $AB$ is a combination of columns of $A$, so $AB$ cannot have a larger column space than $A$. Transposing turns $B$ into the left factor, and transposing does not change rank. So the same argument bounds $\\rank(AB)$ by $\\rank B$.\n",
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