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  "body": "\n## Statement\n\nLet $F$ be a field, let $V$ and $W$ be vector spaces over $F$, and let $T \\colon V \\to W$ be a [[linear map|0NRYHJPQG2]].\nThen $T$ is injective if and only if $\\ker T = \\{0\\}$, where $\\ker T$ is the [[kernel|0S23E6YE83]] of $T$.\n\n## Remarks\n\nThe key step is $T(u) = T(v) \\iff T(u - v) = 0 \\iff u - v \\in \\ker T$.\nThus each nonempty fibre $T^{-1}(w)$ is a coset $v + \\ker T$.\nFor a matrix $A$, the map $x \\mapsto Ax$ is injective if and only if $Ax = 0$ has only the trivial solution.\n\n## Explanations\n\n### L1 human\n\nTo test if a linear map is one to one, we only need to look at the vectors that go to zero.\nIf two vectors have the same image, then their difference goes to zero.\nSo if only $0$ goes to $0$, then no two vectors share an image.\n",
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