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  "body": "\n## Proof\n\nWe use the rank–nullity theorem to compare the kernel and the image.\n\n1. Let $n = \\dim V$. By the [[rank–nullity theorem|0W4FSW1TVB]], $\\dim \\im T + \\dim \\ker T = n$, where $\\im T$ is the [[image|0WSFKKT6FF]] and $\\ker T$ the [[kernel|0S23E6YE83]].\n2. A subspace of [[dimension|03GSZ8RTZJ]] $0$ has the empty family as [[basis|0KCQ9Y3QKA]], so it equals $\\{0\\}$. Conversely $\\{0\\}$ has dimension $0$.\n3. (1 implies 2.) Suppose $T$ is injective. By [[the kernel criterion|0BJBYCGMWE]], $\\ker T = \\{0\\}$, so $\\dim \\ker T = 0$.\n4. By step 1, $\\dim \\im T = n = \\dim V$. Since $\\im T$ is a subspace of $V$, [[a subspace of full dimension is the whole space|0WGAXEK99E]] gives $\\im T = V$. So $T$ is surjective.\n5. (2 implies 1.) Suppose $T$ is surjective. Then $\\im T = V$, so $\\dim \\im T = n$.\n6. By step 1, $\\dim \\ker T = 0$, so $\\ker T = \\{0\\}$ by step 2. By the kernel criterion, $T$ is injective.\n7. (3 is equivalent to 1 and 2.) A map is bijective exactly when it is injective and surjective.\n8. If $T$ is bijective, it is injective, so condition 1 holds. If condition 1 holds, then condition 2 holds by step 4, so $T$ is bijective.\n9. By steps 3 to 8, the three conditions are equivalent.\n\n## Idea\n\nRank–nullity says that the dimensions of the kernel and the image add up to $\\dim V$. A trivial kernel forces the image to have full dimension, and a full image forces the kernel to be trivial. This trade-off needs finite dimension: the shift on sequences is injective but not surjective.\n",
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