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  "body": "\n## Proof\n\nWe factor one block matrix in two ways and compare the determinants.\n\n1. Let $A$ be $m \\times n$ and $B$ be $n \\times m$ over $K$, with $m, n \\ge 1$, since matrices have at least one row and column. Let $M = \\begin{pmatrix} I_m \u0026 -A \\\\ B \u0026 I_n \\end{pmatrix}$, a [[block matrix|01AXG9KZQF]] of size $m + n$.\n2. By the definition of the [[matrix product|076JGP6R3T]], split each sum $\\sum_k X_{rk} Y_{ks}$ into the indices $k \\le m$ and $k \u003e m$. So block matrices with the same splitting multiply block by block.\n3. Let $L_1 = \\begin{pmatrix} I_m \u0026 0 \\\\ B \u0026 I_n \\end{pmatrix}$ and $U_1 = \\begin{pmatrix} I_m \u0026 -A \\\\ 0 \u0026 I_n + BA \\end{pmatrix}$, with [[identity matrices|06DW9BAWB1]] $I_m, I_n$.\n4. By step 2, the blocks of $L_1 U_1$ are $I_m$, $-A$, $B$ and $-BA + I_n + BA = I_n$. So $L_1 U_1 = M$.\n5. Let $U_2 = \\begin{pmatrix} I_m + AB \u0026 -A \\\\ 0 \u0026 I_n \\end{pmatrix}$. We claim that $M = U_2 L_1$.\n6. By step 2, the blocks of $U_2 L_1$ are $(I_m + AB) - AB = I_m$, $-A$, $B$ and $I_n$. So $U_2 L_1 = M$.\n7. The [[transpose|0W809K6AW0]] of $L_1$ is upper block triangular with diagonal blocks $I_m$ and $I_n$.\n8. By the rule for the [[determinant of a transpose|0ZZQ2E1MTE]], the rule for [[block triangular matrices|0J9TVTNB4K]] and the [[determinant of the identity|0Z4BWARC38]], $\\det L_1 = 1$.\n9. By the rule for block triangular matrices, $\\det U_1 = \\det I_m \\cdot \\det(I_n + BA) = \\det(I_n + BA)$.\n10. In the same way, $\\det U_2 = \\det(I_m + AB) \\cdot \\det I_n = \\det(I_m + AB)$.\n11. By the [[product rule|04CPJ6AF00]] and steps 4, 8 and 9, $\\det M = \\det(I_n + BA)$.\n12. By the product rule and steps 6, 8 and 10, $\\det M = \\det(I_m + AB)$.\n13. By steps 11 and 12, $\\det(I_m + AB) = \\det(I_n + BA)$.\n\n## Idea\n\nThe matrix $\\begin{pmatrix} I_m \u0026 -A \\\\ B \u0026 I_n \\end{pmatrix}$ can be made block triangular by clearing either $B$ or $-A$. Clearing $B$ leaves $I_n + BA$ on the diagonal, and clearing $-A$ leaves $I_m + AB$. Both operations keep the determinant, so the two results agree.\n",
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