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  "body": "\n## Proof\n\n1. Let $T \\colon V \\to W$ be a [[linear map|0NRYHJPQG2]]. First, $T(0) = 0$. Indeed, $T(0) = T(0 + 0) = T(0) + T(0)$, and subtracting $T(0)$ gives $T(0) = 0$.\n2. So $0$ lies in the [[kernel|0S23E6YE83]] $\\ker T$, and $\\{0\\} \\subseteq \\ker T$ always holds.\n3. For $u, v \\in V$, linearity gives $T(u) - T(v) = T(u - v)$. So $T(u) = T(v)$ if and only if $u - v \\in \\ker T$.\n4. Suppose $T$ is injective, and let $v \\in \\ker T$. Then $T(v) = 0 = T(0)$ by step 1, so $v = 0$ by injectivity. Hence $\\ker T = \\{0\\}$.\n5. Suppose $\\ker T = \\{0\\}$, and let $u, v \\in V$ with $T(u) = T(v)$. By step 3, $u - v \\in \\ker T$, so $u - v = 0$ and $u = v$. Hence $T$ is injective.\n6. By steps 4 and 5, $T$ is injective if and only if $\\ker T = \\{0\\}$.\n\n## Idea\n\nTwo vectors have the same image exactly when their difference lies in the kernel. So the map can only merge vectors if the kernel holds a nonzero vector.\n",
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  "name": "Proof that a linear map is injective iff its kernel is trivial",
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