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  "body": "\n## Proof\n\nWe factor out $A$ and reduce the rest to a $1 \\times 1$ determinant.\n\n1. Let $A$ be an invertible $n \\times n$ matrix over $K$, and let $u, v \\in K^n$ be column vectors. Let $w = A^{-1} u$.\n2. By [[distributivity|0RXD6EZYBK]] and [[associativity|073E4XC016]], $A (I_n + w v^T) = A + (A A^{-1}) u v^T$.\n3. By the definition of the [[inverse|0TWTS88TWN]], $A A^{-1} = I_n$. So $A (I_n + w v^T) = A + u v^T$, with the [[identity matrix|06DW9BAWB1]] $I_n$.\n4. By the [[product rule|04CPJ6AF00]], $\\det(A + u v^T) = \\det A \\cdot \\det(I_n + w v^T)$.\n5. The matrix $w$ is $n \\times 1$ and $v^T$ is $1 \\times n$. By [[Sylvester's determinant theorem|0G24VQ1088]], $\\det(I_n + w v^T) = \\det(I_1 + v^T w)$.\n6. The matrix $I_1 + v^T w$ is the $1 \\times 1$ matrix with entry $1 + v^T A^{-1} u$.\n7. By the [[Leibniz formula|016PE4VZ8Q]] for $n = 1$, which has one term, a $1 \\times 1$ matrix has its entry as determinant. So $\\det(I_1 + v^T w) = 1 + v^T A^{-1} u$.\n8. By steps 4 to 7, $\\det(A + u v^T) = (1 + v^T A^{-1} u) \\det A$.\n\n## Idea\n\nWrite $A + u v^T = A (I + A^{-1} u v^T)$. The second factor is the identity plus a column times a row. Sylvester's theorem swaps the order of the column and the row, which turns the $n \\times n$ determinant into the number $1 + v^T A^{-1} u$.\n",
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