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  "body": "\n## Statement\n\nLet $A$ be an $n \\times n$ complex matrix with columns $v_1, \\dots, v_n \\in \\C^n$.\nLet $\\norm{v}$ denote the Euclidean [[norm|0V2WP9B1MF]] of $v \\in \\C^n$.\nThen\n$$\\abs{\\det A} \\le \\prod_{i=1}^{n} \\norm{v_i},$$\nwhere $\\det$ is the [[determinant|00KDZRW4C0]].\nEquality holds if and only if some $v_i$ is $0$ or the vectors $v_1, \\dots, v_n$ are pairwise [[orthogonal|0Y2X7P2WB7]].\n\n## Remarks\n\nIf the columns are linearly dependent, the left side is $0$.\nOtherwise the [[Gram–Schmidt process|0VT6P0GNQR]] gives $A = QR$ with $Q$ unitary and $R$ upper triangular with diagonal entries $r_{ii}$.\nThen $\\abs{\\det A} = \\prod_i \\abs{r_{ii}}$, and $\\abs{r_{ii}}$ is the distance from $v_i$ to the span of $v_1, \\dots, v_{i-1}$, so $\\abs{r_{ii}} \\le \\norm{v_i}$.\n\nExample: for $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 -1 \\end{pmatrix}$, the columns are orthogonal with norm $\\sqrt{2}$, and indeed $\\abs{\\det A} = 2 = \\sqrt{2} \\cdot \\sqrt{2}$.\nFor $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$, the bound is $1 \\cdot \\sqrt{2}$, while $\\det A = 1$.\n\nAn equivalent form: if $P$ is an $n \\times n$ [[positive semidefinite|01MMVW8CKC]] matrix, then $\\det P \\le \\prod_i p_{ii}$.\nApply it to $P = A^* A$, whose diagonal entries are $\\norm{v_i}^2$.\nIf all entries of $A$ satisfy $\\abs{a_{ij}} \\le 1$, the inequality gives $\\abs{\\det A} \\le n^{n/2}$, which is the starting point of [[Hadamard's maximal determinant problem|GCH1Z9YFNG]].\n\n## Explanations\n\n### L1 human\n\nThe absolute value of a determinant is the volume of the box spanned by the columns.\nA box with given edge lengths has the largest volume when its edges are at right angles.\nSo the volume is at most the product of the edge lengths, with equality for a rectangular box.\n\n### L2 human\n\nThe inequality bounds the size of a determinant without computing it.\nIt is used to bound the size of integer determinants in algorithms, and it gives the bound $n^{n/2}$ for matrices with entries of absolute value at most $1$.\nThe positive semidefinite form is a special case of Fischer's inequality for block matrices.\n",
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