10. set $ A and B $ to the same-order semi-Definite Matrix, $0 \ Leq s \ Leq 1 $. proof: $ \ Bex \ Sen {A ^ Sb ^ s} _ \ infty \ Leq \ Sen {AB} _ \ infty ^ s. \ EEx $
Proof:
(1 ). first, it is proved that the spectral norm of $ A $ is the maximum singular value of $ A $. in fact, $ \ beex \ Bea \ Sen {A} _ \ infty ^ 2 & =\ Max _ {\ Sen {x} _ 2 = 1} \ Sen {ax} _ 2 ^ 2 \ & =\ Max _ {\ Sen {x} _ 2 = 1} x ^ * a ^ * ax \ & =\ Max _ {\ Sen {x }_ 2 = 1} x ^ * VV ^ * a ^ * u ^ * uavv ^ * x \ & =\ Max _ {\ Sen {y} _ 2 = 1} y ^ * \ diag (S_1 ^ 2, \ cdots, s_p ^ 2) Y \ quad \ sex {Y = V ^ * x} \ & =\ Max _ {\ Sen {y} _ 2 = 1} \ sum _ {I = 1} ^ P s_ I ^ 2 | y_ I | ^ 2 \ & = S_1 ^ 2. \ EEA \ eeex $
(2 ). original Certificate question. $ \ beex \ Bea \ Sen {A ^ Sb ^ s }_\ infty ^ 2 & =\ lm_1 (B ^ SA ^ Sb ^ s) \\& =\ lm_1 (a ^ {2 s} B ^ {2 s}) \\& \ Leq \ SEZ {\ lm_1 (a ^ 2B ^ 2 )} ^ s \ quad \ sex {\ mbox {theorem 3.25 }\\\&=\ SEZ {\ lm_1 (Baab )} ^ s \ & = [\ Sen {AB} _ \ infty ^ 2] ^ s. \ EEA \ eeex $
[Zhan Xiang matrix theory exercise reference] exercise 3.10