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  "body": "\n## Proof\n\nWe first prove that the row rank is at most the column rank for every matrix, and then apply this to the transpose.\n\n1. Let $A = (a_{ik})$ be an $m \\times n$ matrix over $F$. We view its rows as vectors of $F^n$ and its columns as vectors of $F^m$.\n2. Let $\\operatorname{Row}(A) \\subseteq F^n$ be the [[row space|069MCGX635]] and $\\operatorname{Col}(A) \\subseteq F^m$ the [[column space|03AJKXH1AV]]. The spaces $F^n$ and $F^m$ have dimensions $n$ and $m$ by [[the dimension of the coordinate space|0QK7FPE6VE]].\n3. So both spaces are finite-dimensional by the [[dimension of a subspace|0QW084460E]], and they have a [[dimension|03GSZ8RTZJ]]. Let $r = \\dim \\operatorname{Col}(A)$ and $d = \\dim \\operatorname{Row}(A)$.\n4. Claim: $d \\le r$. If $r = 0$, then every column of $A$ is $0$. So $A = 0$, $\\operatorname{Row}(A) = \\{0\\}$, and $d = 0 = r$.\n5. Let $r \\ge 1$, and let $(\\gamma_1, \\dots, \\gamma_r)$ be a [[basis|0KCQ9Y3QKA]] of $\\operatorname{Col}(A)$. Let $C = (c_{il})$ be the $m \\times r$ matrix with columns $\\gamma_1, \\dots, \\gamma_r$.\n6. Each column $\\alpha_k$ of $A$ lies in $\\operatorname{Col}(A)$. So $\\alpha_k = \\sum_{l=1}^r \\rho_{lk} \\gamma_l$ for some scalars $\\rho_{lk} \\in F$. Let $R = (\\rho_{lk})$, an $r \\times n$ matrix.\n7. Comparing entry $i$ in step 6 gives $a_{ik} = \\sum_{l=1}^r c_{il} \\rho_{lk}$ for all $i$ and $k$. By the definition of the [[matrix product|032Q9FB8MT]], this says $A = CR$.\n8. Fix a row index $i$. By step 7, for each $k$, entry $k$ of row $i$ of $A$ is $\\sum_{l=1}^r c_{il} \\rho_{lk}$. So row $i$ of $A$ equals $\\sum_{l=1}^r c_{il} \\cdot (\\text{row } l \\text{ of } R)$.\n9. Let $X \\subseteq F^n$ be the [[span|0DTY8B9HMS]] of the $r$ rows of $R$. By step 8, every row of $A$ lies in $X$, so $\\operatorname{Row}(A) \\subseteq X$.\n10. A basis of $\\operatorname{Row}(A)$ is a linearly independent family of $d$ vectors of $X$. The $r$ rows of $R$ span $X$. By the [[Steinitz exchange lemma|08ZDRN6H41]], $d \\le r$.\n11. This proves the claim $\\dim \\operatorname{Row}(A) \\le \\dim \\operatorname{Col}(A)$ for every matrix $A$.\n12. Now let $A^T$ be the [[transpose|09DXWYRD43]], with entries $(A^T)_{ki} = a_{ik}$. Row $k$ of $A^T$ has entries $a_{1k}, \\dots, a_{mk}$, which is column $k$ of $A$.\n13. So the rows of $A^T$ are the columns of $A$, and the columns of $A^T$ are the rows of $A$. Hence $\\operatorname{Row}(A^T) = \\operatorname{Col}(A)$ and $\\operatorname{Col}(A^T) = \\operatorname{Row}(A)$.\n14. Applying step 11 to $A^T$ gives $\\dim \\operatorname{Col}(A) = \\dim \\operatorname{Row}(A^T) \\le \\dim \\operatorname{Col}(A^T) = \\dim \\operatorname{Row}(A)$.\n15. Together with step 11, $\\dim \\operatorname{Row}(A) = \\dim \\operatorname{Col}(A)$.\n16. By the definition of the [[rank|0ZC9YTXZRY]], $\\rank A^T = \\dim \\operatorname{Col}(A^T) = \\dim \\operatorname{Row}(A) = \\dim \\operatorname{Col}(A) = \\rank A$.\n\n## Idea\n\nChoose a basis of the column space and write every column of $A$ in it; this factors $A = CR$ with $R$ having only $r$ rows. The same factorization shows that every row of $A$ is a combination of those $r$ rows. So the row space is no bigger than the column space, and the transpose gives the reverse inequality.\n",
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