• 2022-05-30
    1) 设[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]为一个[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级实矩阵,且[tex=3.143x1.357]jmW/UUDE3QEpfgsRbhrpUQ==[/tex],证明[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]可以分解成[tex=2.929x1.214]uD+loi5Ndfk9oRNW/S/5NQ==[/tex],其中[tex=0.857x1.214]ChdusW5rAupjge6v/DGHRA==[/tex]是正交矩阵,[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]是一上三角矩阵:[tex=11.929x5.214]iuUhbPg6vGulP+tV2jtZCP8+fINWUOIBYuhILpF13bHy9K9vDpLieRMmpJ2zXt8P5WCwasrT/bhcftZoCydNQZIOF7QAOG8nKDGlYVlVFS54B9tzoOGOGxyZgBkYZKT5OnS6JJpBj7JGFgdTqbS50rB+DFhsxIR915FwxDxWhHkJ5lMjbTLvYpXJ8yVK3iPlHHeABZTdtvP4bnsEbOnI5ErWaTEb143EPJ88etS7vqJ6ismRUFCfZSGkgEeAhnIr[/tex]且[tex=9.071x1.357]o9hjoulZVyyj8haoQFJu/v9xY8hJWIQTjSsa3f/uF9M=[/tex],并证明这个分解是惟一的;2) 设[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级正定矩阵,证明存在一上三角矩阵[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex],使[tex=3.071x1.143]0jLtcygfwX7LHdfUusxcIQ==[/tex].
  • 证:1) 设[tex=9.786x3.5]hB8sGfF8hpZRTKdvt1J/eGgJxZrxpKrca5j28URtYK0lJ/IgklMtJ/FLQpjJUo04yrv1hbfTc6cOPkfCcP159YpIcRMBQ+4/l82k3VkaPKqJ5JSaTFUkSuOOyG7A+uH1D/X64U7ayTs2GOEC3ZXVGDVZEsHgIgjkKFHVmm9gyo4=[/tex]它的[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]个列向量是[tex=5.929x1.0]xJr2ny42kcAcTeyzkoXuGjF5Eh4v7S3Fd052Z/6/FnJsMrtWjHMkU+h8EqXqjCNU[/tex],由于[tex=3.143x1.357]jmW/UUDE3QEpfgsRbhrpUQ==[/tex], 所以[tex=5.929x1.0]xJr2ny42kcAcTeyzkoXuGjF5Eh4v7S3Fd052Z/6/FnJsMrtWjHMkU+h8EqXqjCNU[/tex]是线性无关的,从而是[tex=0.643x1.0]SW0o8G0GHsmLXldwnq7xKg==[/tex]的一组基,现将它直接变成一组标准正交基(同时正交化和单位化),由施密特公式,可导出[tex=22.571x9.929]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[/tex](1)其中,[tex=19.214x5.786]fnpmC2J6JmQBLyo5NmGAz5eRfgnR9bsGFZbabELujeEFufZRvu8dC4MfCLogpzctuoVzLmOpuN+q9RVvNtGDmb/RDKgIcyv8/hWGgYfs0TB5AI3mCa9XiT9L9fuSfl594UxGJkn/js0owyXscuYbNUdnx7VmMPK5+uTfXUBtBRoU0bXlA4xCT0xrnMmtjPZjbZkHw7za/dECydTu/daU/Z2hLhoo/nbCziFt8CXsdz0+l9VolKdx+NJjE1NMxHz7m6yTLkw3GR3iqlgOOx5O9Bm58UgiFYp7bcA+k8nvasSLlwNSiBBiPijVREqftdgKXpxO5+Kqazk+NwBF9naff21YTfMCIpVmQ/L4NeLzUjI=[/tex]那么,[tex=13.786x5.214]fnpmC2J6JmQBLyo5NmGAz0G6YBpo8q9++Hw3ihCqRImxolMMswvg4WHeIFpYYEeLrDcxRYXtrgDIW0BWiHTYeoPnE5ujhkol8MuWnlLYOiXJgwrdrSPVUJsKgk+dhZVe16k6u0pJqm1WcKxNc3luVCP3T963GbcHuu1LQwT+BRl1tIU29vjx/IgFbyhn7MEgojnNv/7LRxLaOb5fnA8bJXQlYAImimfox0sW1J+qmp6aZXK56m7NKqsmOew2zcgGV878Ww8FeyVbsSmasWYo7A==[/tex]其中,[tex=12.786x1.357]pn8yhbOgFnlSiaZzhkiMcbVg8CpRYGQLQvPbv/VmreoeJhlKtb6xFKZRN3ur9zUnuIQkstfKD8JtikDTRY3KZA==[/tex]这就是说,[tex=15.857x1.357]00BnaYlD8t/B0po9R6JDsMNA1VZ3Zj43R5vE0MpfxH4jqHIm0ruQNAAWfmq/DsLQNQHBlpVZeqe+LMgCD0rzjK3B4z/f/PPukOkUPbxpkMltx1RX3DzRSCDx0RsZzJawsp/+IPiJowwK6PqXwxndvw==[/tex][tex=10.429x5.214]dEdrC9SQsN/3Vx39SaFo4LU9Y7DIuXtLbDrA/rVryJXQbplHpggBmNENtsWwKp5rkWSIXnNXZEjw2vX7/iBKz0gwRmzk2AJX7c3BbiYHdnsnv6rsczggRbmHjrvsAT61p8Ch46fKT97zY2kbMPgaVbEyZRZd0HOvFTI4mKrc9Mz7t/t/N8XpB07ktvaHqJbYLk2zDb5aQpa85O0AJeUcjQJZzUzQa+NfE/bqn0pylQQ=[/tex](2)令[tex=9.429x3.929]iuUhbPg6vGulP+tV2jtZCCkRNgnA4LWedvmVp2jqKpxlGDsL3FrMflxVdvQ0NJglYLmy6A3j+PrzpOudaVUOqxofE8vy358bt31arUt4AD4kPMCB0yEwzgEFEB0oCkGFtN9GavUrok9zuolmDs4W37oRGflKzV2BDxAY2CumPuU=[/tex],那么[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]是上三角阵,且主对角线元素[tex=2.714x1.214]CiWWmu88v9CJxMNcf0HyjQ==[/tex].另外,由式(1)可以看出[tex=0.786x1.0]0uModCqSmC8cnK6eUTOQkQ==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]维列向量,不妨记为[tex=12.714x5.357]M8JPsen0IAHVUOMK/iwrJ96PfHajLto70rFEL3s7P54yY0hdlQ+2VK+yHITqKsPkc7t099teeXC6v0EiKSlkVrUK/xstsmg19jW6s9QHDHCVpjxh7zgFW7m4OwKOoFodKZxK9LoMkUWI/BsGyjqRZDcwVfj+jZzC7ca8WpR/Ds4=[/tex]且令,[tex=15.143x4.071]5ZuEj1KRR/p9rD5ciF5Q2iL6zlwsedOohj0aTKX1CQGz3xGrt9TQeaRaTh3gw7+jFBOyCSIGNTP1wCO5OQPY/Lp7kp0z7zw3dZZZglTdIFpzDwlsFQlfZ5VqkJV5yrXo1x6UG5/jwf7LUvCbH5FsZxXa1JbdI3NhGWz+GQ2cdMBM54tAwX37n7tNa03WIaucpPmiQc+NjdhpA4SqWSgjbQ==[/tex]那么代入式(2)后有[tex=2.929x1.214]ZfxJvjEhWh7Ak1fExa4DgQ==[/tex]由于[tex=5.5x1.0]EdkZrZloMH3+0+6SaxtaRFUP8pTZ8X33o7s2iHbn/faiguHLEv4qSHn5rWTUamng[/tex]是标准正交基,故[tex=0.857x1.214]ChdusW5rAupjge6v/DGHRA==[/tex]是正交矩阵.再证惟一性.设[tex=5.786x1.214]2WVUTjYe5OlcfgUPlODBUhNKK6DyXAMjdK2VJSec5uM=[/tex](3)是两种分解,其中[tex=2.429x1.214]YWTUzACzuxVEydo2jA2oxA==[/tex] 是正交阵,[tex=2.0x1.214]HjtwNLCrdjJYh2BDXQXxkQ==[/tex] 是主对角线元素大于0的上三角阵, 那么由式(3),得[tex=5.929x1.5]SQ7ScS1QszM5pE93lMMFzhmI+frpjqXoaMFhvGenK9U=[/tex]由于[tex=2.571x1.5]A/jFKSsSaOCxY1vmd4Zfkw==[/tex]是正交阵,从而[tex=2.643x1.429]jvUigWffyCmG60vwzPxlmOq5DJ1M3knqdmI9KjNPCmc=[/tex]也为正交阵,且为上三角阵,那么由本章习题第13题知,[tex=2.643x1.429]jvUigWffyCmG60vwzPxlmOq5DJ1M3knqdmI9KjNPCmc=[/tex]是主对角线元素为1 或-1 的对角阵,但是[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]与[tex=1.0x1.214]IwnIX+ymTn2PT96bYYErrQ==[/tex] 主对角线元素为正,所以[tex=2.643x1.429]jvUigWffyCmG60vwzPxlmOq5DJ1M3knqdmI9KjNPCmc=[/tex]的主对角线上的元素只能是1,换句话说,就是[tex=4.143x1.429]3wM7yTG/Xaf9SXyzyX39FA==[/tex],即[tex=2.357x1.214]NCnsvEGN6NYCH0xfjRoT/A==[/tex]再由[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]是非退化及式(3),所以[tex=2.714x1.214]L9Zyj+FnnAxCV0itjTwc7g==[/tex]从而分解是惟一的.2) 因为[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是正定的,由第五章知[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]与[tex=0.786x1.0]XvHgf70VtK2FH5G93l0k3g==[/tex]合同,即存在可逆阵[tex=0.714x1.0]J/aA9EEo0KmJFnWWfX7LmQ==[/tex],使          [tex=6.571x1.143]grANrPz3o+ZIT/xxH8+axd/IXOetBsmtnjK2K2IFPdU=[/tex](4)再由上面证明,知[tex=2.929x1.214]IYuHrK4uqvAdvSb41i8rnA==[/tex],其中[tex=0.857x1.214]ChdusW5rAupjge6v/DGHRA==[/tex]为正交阵,[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]为上三角阵,再将[tex=0.714x1.0]J/aA9EEo0KmJFnWWfX7LmQ==[/tex]代入式(4),即有         [tex=7.214x1.357]R9YsoAtxNH+2CcCSWbdsjUi5jqid73QMGagNN+8Lpt71+7g3Hvh/8BxFaQQhMwuu[/tex]

    举一反三

    内容

    • 0

      证明:如果[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级正定矩阵,[tex=0.786x1.0]ri6gmnf1+J9dGqG5/1sV6A==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级半正定矩阵,则[tex=2.786x1.143]7OI9Dpqsob5Abz33m0rKpw==[/tex]是正定矩阵.

    • 1

      证明:如果[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级正定矩阵,[tex=0.786x1.0]ri6gmnf1+J9dGqG5/1sV6A==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级半正定矩阵,那么[tex=2.286x1.143]7OI9Dpqsob5Abz33m0rKpw==[/tex]是正定矩阵。

    • 2

      主对角线上全是1 的上三角矩阵称为特殊上三角矩阵.1) 设[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是一对称矩阵,[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]为特殊上三角矩阵,而[tex=3.857x1.143]qPbOPa2llKVdw9JpP1uKuIR3ci6q7f1T05yD8ZrWl+k=[/tex],证明:[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]与[tex=0.786x1.0]ri6gmnf1+J9dGqG5/1sV6A==[/tex]的对应顺序主子式有相同的值;2) 利用以上结果证明定理7的充分性.

    • 3

      设[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是一个[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级正交矩阵,证明:如果[tex=3.429x1.357]KfxiXgR+wZCad+SOlQefBQ==[/tex],那么-1是[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]得一个特征值。

    • 4

      设[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是[tex=2.357x1.071]QArHY/B/HPaeI4OFb8f5sA==[/tex]矩阵.证明:[tex=0.786x1.0]Yn3GgEZev6SOu2r4v1WnCw==[/tex]是行满秩矩阵当且仅当存在[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]级可逆矩阵[tex=0.857x1.214]ChdusW5rAupjge6v/DGHRA==[/tex],使得[tex=4.643x1.357]RqsCL9/WuCBwDvrRjd44OA==[/tex].