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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $A$ and $B$ be $n \\times n$ [[Hermitian matrices|0P17NGQQMQ]].\nThen $A \\succeq B$ in the *Loewner order* if $A - B$ is [[positive semidefinite|01MMVW8CKC]].\nAlso $A \\succ B$ if $A - B$ is [[positive definite|0KX66CZMBB]].\n\n## Remarks\n\nThe relation $\\succeq$ is a [[partial order|0QV2JS9RJ3]].\nIt is transitive because sums of positive semidefinite matrices are positive semidefinite.\nIt is antisymmetric because a matrix that is both positive semidefinite and negative semidefinite is $0$.\nExample: $\\begin{pmatrix} 2 \u0026 1 \\\\ 1 \u0026 2 \\end{pmatrix} \\succeq I$, because the difference $\\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ has eigenvalues $2$ and $0$.\nNon-example: $\\operatorname{diag}(1, 0) \\not\\succeq \\operatorname{diag}(0, 1)$, because the difference $\\operatorname{diag}(1, -1)$ has the eigenvalue $-1$; the converse fails too, so the order is not total.\nThe order does not respect squares: for $A = \\begin{pmatrix} 2 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 0 \\end{pmatrix}$, $A \\succeq B$, but $A^2 - B^2 = \\begin{pmatrix} 4 \u0026 3 \\\\ 3 \u0026 2 \\end{pmatrix}$ has determinant $-1$.\nOn the other hand, $A \\succeq B \\succeq 0$ implies $A^{1/2} \\succeq B^{1/2}$.\n",
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