证明定理8.5.1.
设[tex=6.5x1.357]v+YpN3YF9z9lZ7BsT3xcCniyIGKjzNxF4HdcgkjPrzdBT5e53PoE7fhc3knpIhRE6p26xFHDz8a9uo47Lf2TAQ==[/tex]和[tex=6.286x1.357]7cPUqIFMOXpGOd0aJs6Mb4F3YpGHemt1bw5I7O9SvvMAL6Ed5IzNTM0+2FkU82sdt8rMlvNlAk85jFKQQb+/Tw==[/tex]是[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]维酉空间[tex=0.643x1.0]SW0o8G0GHsmLXldwnq7xKg==[/tex]的两个规范正交基,而[tex=4.929x1.5]xEg3s7+tlOn8nT2cGb4VVEkgwybXEtNwKWaINUQaKcE2XDF16Tg/UfylXUHKYNer[/tex]是过渡矩阵,那么[tex=10.214x3.286]EZQZk3jkuD5TM0rjTloPlE/A/+CU2d8n/0/Gk/kX4xOI0oew681zEhGT6Xce1rS0ib85wlmo0v78ijQ4EIZSZkLetE3Nyol/0jKsg7WH4Ho=[/tex]且[tex=10.0x3.357]/N4VjDQHvWDt28l0trOmudkhcJ45dnx1/XJg9oIYdX/QEIFSThawXEpoofDUh+aFeMRknc0/gXSlGtQcbGmPA51fGSMrxr7AwI06CcznJa1vafpGju6SrB93n5/M/cQrIUnGmANpmN/2Xa+sgSlm52lHpcZ5exJ+owNOnJH9O3U=[/tex]另一方面,由于[tex=6.5x1.357]v+YpN3YF9z9lZ7BsT3xcCniyIGKjzNxF4HdcgkjPrzdBT5e53PoE7fhc3knpIhRE6p26xFHDz8a9uo47Lf2TAQ==[/tex]也是规范正交基,所以[tex=27.214x3.357]/N4VjDQHvWDt28l0trOmudkhcJ45dnx1/XJg9oIYdX/QEIFSThawXEpoofDUh+aFwYeT9EnfleEee9TVoYCx1FIum/Fa/k4hDclzRA6nt+LwnCf18AqCNlx+24NhvvTJC6GWAmuIip1vxxtZvt34xIfomG/fYlJBn4zu2lra0p0BAXRLoK4lg/I6YPvUkk/Dc3vERqwv2ZEp++IH6xyzU70S+F+5VmYaq4l7zqhsB7fR3IxZDLrhzApo9JQF76eyeVrhLhY+VQ6VfG1dm0zjkng510tvRz5AuqqZzQnFkwX6usa2qzTOzRcM9rQPvlzAlrXrUs1YrYzLtvalN0jaHiWQ6gmmEUP7oj+SSr22+22SDOwvtNwfAeqtPxCDG6fm[/tex]从而[tex=11.429x3.357]qrSlL2RK8Ra3t/EJQhFzm2bPKiICPVW3q1K2wsWUTx1MnWGQ4JwsrR5S4WTOvfnX2xr7Xl+P16O20A7aABukf3gATASMO/hifCwUIv7qdlSJb3DIAXMfkkJrbvKVkAgLQB0IRAON3jWSyyasIMjA+w==[/tex],即[tex=5.643x1.429]CbGtS09tH2Huz1ChLHqivssfrqG4R2Ib6P93hdnXfG1TqgxbZlp0NnOLL/S8sJK8pg0kb+jxmaYqafsjYps3Vg==[/tex]从而定理 8. 5. 1 得证,即[tex=0.643x0.786]SBMIs+VUk7//BOpfqlQl0w==[/tex]维酉空间一个规范正交基到另一个规范正交基的过渡矩阵是酉矩阵.
举一反三
内容
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