[Zhan Xiang matrix theory exercise reference] Exercise 5.2

2. $ \ im A $ indicates the image space of $ A \ in M_n $: $ \ bex \ im A =\sed {Ax; x \ in \ bbC ^ n }. \ eex $ Set $ A, B \ in M_n $ as orthogonal projection matrix, satisfying $ \ bex \ sen {A-B} _ \ infty <1. \ eex $ proof: $ \ bex \ dim \ im A = \ dim \ im B. \ eex $

 

 

Proof: use the reverse verification method. set $ \ dim \ im A <\ dim \ im B $, then $ \ beex \ bea n & =\ dim \ im A ++ \ dim \ ker A \\& <\ dim \ im B ++ \ dim \ ker A \\& = \ dim (\ im B + \ ker) + \ dim (\ im B \ cap \ ker A) \ & \ leq n + \ dim (\ im B \ cap \ ker ). \ eea \ eeex $ So, $ \ bex \ dim (\ im B \ cap \ ker A)> 0, \ eex $ \ bex \ exists \ 0 \ neq y \ in \ bbC ^ n, \ st y = Bx, \ quad Ay = 0. \ eex $ and $ \ bex (A-B) y =-By =-B (Bx) =-B ^ 2x =-Bx =-y. \ eex $ this description $-1 $ is an feature value for $ A-B $, $ \ bex \ sen {A-B }_\ infty = \ max_ I | \ lm_ I (A-B) | = 1. \ eex $ this is a conflict. therefore, there is a conclusion.

[Zhan Xiang matrix theory exercise reference] Exercise 5.2