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  "body": "\n## Statement\n\nLet $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, and let $T \\colon V \\to V$ be a linear map.\nThen these conditions are equivalent:\n\n1. $T$ is injective.\n2. $T$ is surjective.\n3. $T$ is bijective.\n\n## Remarks\n\nThe proof uses the [[rank–nullity theorem|0W4FSW1TVB]]: $\\ker T = 0$ if and only if $\\dim \\im T = \\dim V$.\nIt fails in infinite dimensions: on $F[x]$, multiplication by $x$ is injective and not surjective.\nIn characteristic $0$, the derivative on $F[x]$ is surjective and not injective.\nFor square matrices, this is the statement that $Ax = 0$ has only the trivial solution if and only if $Ax = b$ has a solution for each $b$.\n\n## Explanations\n\n### L1 human\n\nFor a linear map from a finite-dimensional space to itself, one to one and onto are the same thing.\nIf no information is lost, then nothing in the target can be missed, because the dimensions match.\nThis is like a map from a finite set to itself.\n",
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