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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ be an $m \\times n$ matrix over $F$.\nThen the dimension of the [[row space|069MCGX635]] of $A$ equals the dimension of the [[column space|03AJKXH1AV]] of $A$.\nEquivalently, $\\rank A^T = \\rank A$, where $A^T$ is the [[transpose|09DXWYRD43]].\n\n## Remarks\n\nOne proof uses row reduction: row operations keep the row space and the dimension of the column space.\nIn reduced row echelon form, both dimensions equal the number of pivots.\nAnother proof uses the [[annihilator|0ASHNYF7AG]]: the dual map has the transposed matrix and the same rank.\nOver $\\C$, also $\\rank A^* = \\rank A$.\n\n## Explanations\n\n### L1 human\n\nThe number of independent rows of a matrix always equals the number of independent columns.\nThis is surprising, because rows and columns live in different spaces.\nFor a $3 \\times 5$ matrix, it shows at once that the rank is at most $3$.\n",
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