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      "label": "> [!theorem] <a class=\"object-permalink\" href=\"#theorem-5-3\">定理 5.3</a> · 西尔维斯特惯性定律 (Sylvester’s law of inertia)",
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        "quote": "\n\n> [!theorem] <a class=\"object-permalink\" href=\"#theorem-5-3\">定理 5.3</a> · 西尔维斯特惯性定律 (Sylvester’s law of inertia)\n>\n> (西尔维斯特) 在 $\\eta _ { i }$ 中，$+$ 与 $-$ 符号的数目与标准正交基的选取无关。\n\n> [!proof]- 证明\n>\n> 设 $\\{ e _ { i } \\}$ 与 $\\{ f _ { j } \\}$ 为两个标准正交基，使得\n>\n> $$\n> e _ {1} \\cdot e _ {1} = \\dots = e _ {r} \\cdot e _ {r} = + 1, \\qquad e _ {r + 1} \\cdot e _ {r + 1} = \\dots = e _ {n} \\cdot e _ {n} = - 1,\n> $$\n>\n> $$\n> f _ {1} \\cdot f _ {1} = \\dots = f _ {s} \\cdot f _ {s} = + 1, \\quad f _ {s + 1} \\cdot f _ {s + 1} = \\dots = f _ {n} \\cdot f _ {n} = - 1.\n> $$\n>\n> 若 $s > r$，则向量 $f _ { 1 } , \\ldots , f _ { s }$ 与 $e _ { r + 1 } , \\ldots , e _ { n }$ 构成一组 $s + n - r > n = \\dim V$ 个向量，且它们之间必存在一个非平凡线性关系，\n>\n> $$\n> a ^ {1} f _ {1} + \\dots + a ^ {s} f _ {s} + b ^ {1} e _ {r + 1} + \\dots + b ^ {n - r} e _ {n} = 0.\n> $$\n>\n> 由于$a ^ { i }$不可能全部为零，因为$e _ { i }$构成线性无关集。类似地，并非所有$b ^ { j }$都会为零。令\n>\n> $$\n> u = a ^ {1} f _ {1} + \\dots + a ^ {s} f _ {s} = - b ^ {1} e _ {r + 1} - \\dots - b ^ {n - r} e _ {n} \\neq 0\n> $$\n>\n> 我们得到了矛盾\n>\n> $$\n> u \\cdot u = \\sum_ {i = 1} ^ {s} (a ^ {i}) ^ {2} > 0 \\quad \\text { and } \\quad u \\cdot u = - \\sum_ {j = 1} ^ {n - r} (b ^ {j}) ^ {2} <   0.\n> $$\n>\n> 因此，$r = s$ 与这两个基必须具有完全相同数量的 $+$ 和 $-$ 符号。\n",
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        "quote": "\n\n> [!theorem] <a class=\"object-permalink\" href=\"#theorem-5-3\">Theorem 5.3</a> · Sylvester’s law of inertia\n>\n> (Sylvester) The number of $+$ and $-$ signs among the $\\eta _ { i }$ is independent of the choice of orthonormal basis.\n\n> [!proof]- Proof\n>\n> Let $\\{ e _ { i } \\}$ and $\\{ f _ { j } \\}$ be two orthonormal bases such that\n>\n> $$\n> e _ {1} \\cdot e _ {1} = \\dots = e _ {r} \\cdot e _ {r} = + 1, \\qquad e _ {r + 1} \\cdot e _ {r + 1} = \\dots = e _ {n} \\cdot e _ {n} = - 1,\n> $$\n>\n> $$\n> f _ {1} \\cdot f _ {1} = \\dots = f _ {s} \\cdot f _ {s} = + 1, \\quad f _ {s + 1} \\cdot f _ {s + 1} = \\dots = f _ {n} \\cdot f _ {n} = - 1.\n> $$\n>\n> If $s > r$ then the vectors $f _ { 1 } , \\ldots , f _ { s }$ and $e _ { r + 1 } , \\ldots , e _ { n }$ are a set of $s + n - r > n = \\dim V$ vectors and there must be a non-trivial linear relation between them,\n>\n> $$\n> a ^ {1} f _ {1} + \\dots + a ^ {s} f _ {s} + b ^ {1} e _ {r + 1} + \\dots + b ^ {n - r} e _ {n} = 0.\n> $$\n>\n> The $a ^ { i }$ cannot all vanish since the $e _ { i }$ form an l.i. set. Similarly, not all the $b ^ { j }$ will vanish. Setting\n>\n> $$\n> u = a ^ {1} f _ {1} + \\dots + a ^ {s} f _ {s} = - b ^ {1} e _ {r + 1} - \\dots - b ^ {n - r} e _ {n} \\neq 0\n> $$\n>\n> we have the contradiction\n>\n> $$\n> u \\cdot u = \\sum_ {i = 1} ^ {s} (a ^ {i}) ^ {2} > 0 \\quad \\text { and } \\quad u \\cdot u = - \\sum_ {j = 1} ^ {n - r} (b ^ {j}) ^ {2} <   0.\n> $$\n>\n> Hence $r = s$ and the two bases must have exactly the same number of $+$ and $-$ signs.\n",
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