1. $ A \ in M_n $ is called an orthogonal projection matrix. If $ A $ is A Hermite matrix and its idempotence is: $ \ bex A ^ * = A ^ 2. \ eex $ Proof: If $ A, B \ in M_n $ is an orthogonal projection matrix, then $ \ sen {A-B} _ \ infty \ leq 1 $.
Proof: By $ A ^ * = A $ Zhi $ A $, you can perform the right-to-right corner. the feature value of $ A ^ 2 = A $ Zhi $ A $ is $0 $ or $1 $. therefore, $ Makes $ \ bex A = U ^ * \ diag (I _r, 0) U, \ quad r = \ rank (). \ eex $, $ \ beex \ bea x ^ * Ax & = x ^ * U ^ * \ diag (I _r, 0) ux \ & = y ^ * \ diag (I _r, 0) y \ quad \ sex {y = Ux }\\\& =\ sum _ {I = 1} ^ r | y_ I | ^ 2 \\\& \ in \ sez {0, \ sen {x} ^ 2 }. \ eea \ eeex $ similarly, $ \ bex x ^ * Bx \ in \ sez {0, \ sen {x} ^ 2 }. \ eex $ Because $ A-B $ is Hermite array, $ \ bex \ sen {A-B} _ \ infty = s_1 (A-B) = \ max_ I | \ lm_ I (A-B) |. \ eex $ Set $ \ lm $ to any feature value of $ A-B $, $0 \ neq x \ in \ bbC ^ n $ to its feature vector, then $ \ beex \ bea (A-B) x & =\ lm x, \ | \ lm | & =\ frac {| x ^ * (A-B) x |}{\ sen {x} ^ 2 }\\\&=\ sev {\ frac {x ^ * Ax }{\ sen {x} ^ 2}-\ frac {x ^ * Bx} {\ sen {x} ^ 2 }\\\& \ in \ sez {0, 1 }, \ eea \ eeex $ the last step is because $ \ bex 0 \ leq s \ leq 1, \ quad 0 \ leq t \ leq 1 \ ra-1 \ leq s-t \ leq 1 \ ra | s-t | \ leq 1. \ eex $ hence $ \ bex \ sen {A-B} _ \ infty = \ max_ I | \ lm_ I (A-B) | \ leq 1. \ eex $
[Zhan Xiang matrix theory exercise reference] exercise 5.1