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  "body": "\n## Definition\n\nLet $F$ be a field and let $p, q \\ge 1$.\nLet $M$ be a $(p + q) \\times (p + q)$ [[block matrix|0ASDFEQS6P]] over $F$,\n$$M = \\begin{pmatrix} A \u0026 B \\\\ C \u0026 D \\end{pmatrix},$$\nwhere $A$ is $p \\times p$, $B$ is $p \\times q$, $C$ is $q \\times p$ and $D$ is $q \\times q$.\nIf $A$ is [[invertible|0EB9N0S33C]], the *Schur complement of $A$ in $M$* is the $q \\times q$ matrix\n$$M/A = D - C A^{-1} B.$$\nIf $D$ is invertible, the *Schur complement of $D$ in $M$* is the $p \\times p$ matrix\n$$M/D = A - B D^{-1} C.$$\n\n## Remarks\n\nExample: let $M = \\begin{pmatrix} 2 \u0026 1 \\\\ 4 \u0026 5 \\end{pmatrix}$ with $p = q = 1$ and $A = (2)$.\nThen $M/A = 5 - 4 \\cdot \\tfrac{1}{2} \\cdot 1 = 3$, and indeed $\\det M = 10 - 4 = 6 = 2 \\cdot 3$.\nNon-example: for $M = \\begin{pmatrix} 0 \u0026 1 \\\\ 1 \u0026 0 \\end{pmatrix}$ and $A = (0)$, the block $A$ is not invertible, so $M/A$ is not defined, although $M$ itself is invertible.\nThe definition comes from the factorisation\n$$M = \\begin{pmatrix} I_p \u0026 0 \\\\ C A^{-1} \u0026 I_q \\end{pmatrix} \\begin{pmatrix} A \u0026 0 \\\\ 0 \u0026 M/A \\end{pmatrix} \\begin{pmatrix} I_p \u0026 A^{-1} B \\\\ 0 \u0026 I_q \\end{pmatrix},$$\nwhich gives $\\det M = \\det A \\cdot \\det(M/A)$; see [[determinant via the Schur complement|0KA7A4ENVJ]].\nIn particular, when $A$ is invertible, $M$ is invertible if and only if $M/A$ is invertible.\nThe same formulas hold over any commutative ring when $A$ is invertible.\n",
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