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  "body": "\n## Proof\n\nLet $T$ be a [[linear map|0NRYHJPQG2]] as in the statement. We extend a basis of the kernel to a basis of $V$. Then we show that the new vectors map to a basis of the image.\n\n1. Let $n = \\dim V$. The [[kernel|0S23E6YE83]] $\\ker T$ is a [[subspace|00YC1D652K]] of $V$. It contains $0$, since $T(0) = T(0 + 0) = 2T(0)$ gives $T(0) = 0$.\n2. If $u, v \\in \\ker T$ and $a \\in F$, then $T(u + v) = T(u) + T(v) = 0$ and $T(av) = a T(v) = 0$. So $\\ker T$ is closed under both operations.\n3. The [[image|0WSFKKT6FF]] $\\im T$ is a subspace of $W$. It contains $T(0) = 0$, and $T(u) + T(v) = T(u + v)$ and $a T(v) = T(av)$ lie in $\\im T$.\n4. By the [[dimension of a subspace|0QW084460E]], $\\ker T$ is finite-dimensional. Let $k = \\dim \\ker T$ and let $(v_1, \\dots, v_k)$ be a [[basis|0KCQ9Y3QKA]] of $\\ker T$.\n5. This family is [[linearly independent|0CRMGPAH83]] in $V$. By [[extension to a basis|0M4SXE5WW3]] and the [[dimension theorem|0WK0Y8RH1M]], there is a basis $(v_1, \\dots, v_k, u_1, \\dots, u_{n-k})$ of $V$.\n6. The vectors $T(u_1), \\dots, T(u_{n-k})$ [[span|088RYKAWK9]] $\\im T$. Let $w \\in \\im T$, say $w = T(v)$ with $v \\in V$.\n7. Write $v = \\sum_{i=1}^k a_i v_i + \\sum_{j=1}^{n-k} c_j u_j$. Since $T(v_i) = 0$, linearity gives $w = T(v) = \\sum_{j=1}^{n-k} c_j T(u_j)$.\n8. The vectors $T(u_1), \\dots, T(u_{n-k})$ are linearly independent. Suppose $\\sum_{j=1}^{n-k} c_j T(u_j) = 0$ with $c_j \\in F$.\n9. By linearity, $T\\left(\\sum_j c_j u_j\\right) = 0$, so $\\sum_j c_j u_j \\in \\ker T$. Since $(v_i)$ spans $\\ker T$, $\\sum_j c_j u_j = \\sum_i a_i v_i$ for some $a_i \\in F$.\n10. Then $\\sum_i a_i v_i - \\sum_j c_j u_j = 0$. The basis of step 5 is linearly independent, so all $c_j = 0$.\n11. By steps 6 to 10, $(T(u_1), \\dots, T(u_{n-k}))$ is a basis of $\\im T$. So $\\dim \\im T = n - k$, by the definition of [[dimension|03GSZ8RTZJ]].\n12. Hence $\\dim \\im T + \\dim \\ker T = (n - k) + k = n = \\dim V$. In other words, the [[rank|0ZZXSBTR4J]] plus the [[nullity|0APCFW7639]] of $T$ equals $\\dim V$.\n\n## Idea\n\nSplit a basis of $V$ into a basis of the kernel and some extra vectors. The kernel part is sent to $0$. The extra vectors are sent to a basis of the image, because any relation among their images would put a combination of them into the kernel.\n",
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