[Zhan Xiang matrix theory exercise reference] exercise 4.17

17. (ando-Zhan) is set to $ A, B \ In M_n $ semi-definite, $ \ Sen {\ cdot} $ is an undo norm, then $ \ Bex \ Sen {(a + B) ^ r} \ Leq \ Sen {A ^ r + B ^ r }, \ Quad (0 <r \ Leq 1), \ EEx $ \ Bex \ Sen {(A + B) ^ r} \ geq \ Sen {A ^ r + B ^ r}, \ Quad (1 \ Leq r <\ infty ). \ EEx $

 

Proof: (1 ). verify when $0 <r \ Leq 1 $, $ \ Bex \ Sen {(A + B) ^ r} \ Leq \ Sen {A ^ r + B ^ r }. \ EEx $ is based on the fan principle. You only need to verify $ \ Bex S (a + B) ^ r) \ prec_w S (a ^ r + B ^ r ). \ EEx $ for $ \ forall \ 1 \ Leq k \ Leq N $, $ \ Bex \ sum _ {I = 1} ^ k \ lm_ I (a + B) ^ r) \ Leq \ sum _ {I = 1} ^ k \ lm_ I (a ^ r + B ^ r ). \ EEx $ is the proof of chapter 3 and question 11th, which is equivalent to proof: $ \ Bex \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {(a + B) ^ rx_ I, x_ I} \ Leq \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {(a ^ r + B ^ r) X_ I, X_ I }. \ EEx $ $ F (t) = t ^ r$ is an operator Monotonic function on $ [0, \ infty) $ with Integral Representation (4.13) $ \ Bex t ^ r = \ Al + \ beta T + \ int_0 ^ \ infty \ frac {st} {S + t} \ RD \ Mu (s ), \ quad \ beta \ geq 0. \ EEx $ we only need to prove that for $ S> 0 $, $ \ beex \ Bea \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {S (A + B) (Si + A + B) ^ {-1} X_ I, x_ I} & \ Leq \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {SA (Si + a) ^ {-1} X_ I, x_ I }\\& \ quad + \ sum _ {I = 1} ^ k \ SEF {Sb (Si + B) ^ {-1} X_ I, X_ I }. \ EEA \ eeex $ Replace $ A/S with $ A and B $ respectively, $ B/S $, that is, to prove $ \ beex \ Bea \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {(A + B) (I + A + B) ^ {-1} X_ I, x_ I} & \ Leq \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {A (I + a) ^ {-1} X_ I, x_ I }\\& \ quad + \ Max _ {\sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {B (I + B) ^ {-1} X_ I, X_ I }. \ EEA \ eeex $ select $ X_ I $ as the unit feature vector from the largest feature value of $ I $ under $ A + B $, that is, to prove $ \ Bex \ sum _ {I = 1} ^ k \ SEF {(A + B) (I + A + B) ^ {-1} X_ I, x_ I} \ Leq \ sum _ {I = 1} ^ k \ SEF {A (I + a) ^ {-1} X_ I, x_ I} + \ sum _ {I = 1} ^ k \ SEF {B (I + B) ^ {-1} X_ I, X_ I }. \ EEx $ note $ c = (I + A + B) ^ {-\ frac {1} {2 }}$, then $ \ BEX (A + B) (I + A + B) ^ {-1} = (a + B) C ^ 2 = C (a + B) C = CAC + CBC. \ EEx $ it must prove that $ \ bee \ label {4_17_a} \ sum _ {I = 1} ^ k \ SEF {cacx_ I, x_ I} \ Leq \ sum _ {I = 1} ^ k \ SEF {A (I + a) ^ {-1} X_ I, X_ I }, \ EEE $ \ bee \ label {4_17_ B} \ sum _ {I = 1} ^ k \ SEF {cbcx_ I, x_ I} \ Leq \ sum _ {I = 1} ^ k \ SEF {B (I + B) ^ {-1} X_ I, X_ I }. \ EEE $ the two proofs are the same, so only \ eqref {4_17_a} must be proved as follows. by $ \ beex \ BEA (a + B) X_ I = \ lm_ix_ I & \ rA (I + A + B) X_ I = (\ lm_ I + 1) x_ I \ & \ rA C ^ 2x_ I = \ frac {1} {1 + \ lm_ I} X_ I \ & \ rA cx_ I = \ frac {1} {\ SQRT {1 + \ lm_ I} X_ I \ EEA \ eeex $ $ X_ I $ is the unit feature vector from the small feature value $ \ mu_ I $ of $ C $, $ I = 1, \ cdots, K $. order $ \ Bex U_1 = (x_1, \ cdots, X_k ), \ EEx $ to prove $ \ bee \ label {4_17_tr} \ tr \ SEZ {U_1 ^ * cacu_1} \ Leq \ tr \ SEZ {U_1 ^ * a (a + i) ^ {-1} U_1 }. \ EEE $ set $ X_ I $ as defined above, $ I = k + 1, \ cdots, N $, $ \ Bex u_2 = (X _ {k + 1 }, \ cdots, X_n), \ quad u = (U_1, u_2), \ EEx $ then $ \ Bex cu_1 = u_1d_1, \ quad Cu_2 = u_2d_2, \ quad U \ mbox {}, \ EEx $ \ Bex d_1 = \ diag (\ mu_1, \ cdots, \ mu_k ), \ quad D_2 = \ diag (\ Mu _ {k + 1}, \ cdots, \ mu_n), \ quad \ mu_1 \ Leq \ cdots \ Leq \ mu_n. \ EEx $, $ \ beex \ Bea \ tr \ SEZ {U_1 ^ * cacu_1} & =\ Sen {A ^ \ frac {1} {2} cu_1} _ f ^ 2 \\& = \ Sen {u ^ * a ^ \ frac {1} {2} u_1d_1} _ f ^ 2 \ & =\ Sen {\ sex {U_1 ^ * a ^ \ frac {1} {2} u_1d_1 \ atop u_2 ^ * a ^ \ frac {1} {2} u_1d_1} _ f ^ 2 \ & = \ Sen {U_1 ^ * ^ \ frac {1} {2} u_1d_1} _ f ^ 2 + \ Sen {u_2 ^ * a ^ \ frac {1} {2} u_1d_1} _ f ^ 2 \\ & \ Leq \ mu_k ^ 2 \ Sen {U_1 ^ * a ^ \ frac {1} {2} u_1d_1} _ f ^ 2 + \ Sen {u_2 ^ * a ^ \ frac {1} {2} U_1} _ f ^ 2; \ tr \ SEZ {U_1 ^ * a (a + I) ^ {-1} U_1} & =\ tr \ SEZ {U_1 ^ * a ^ \ frac {1} {2} (a + I) ^ {-1} a ^ \ frac {1} {2} U_1 }\\\& \ geq \ tr \ SEZ {U_1 ^ * a ^ \ frac {1} {2} CA ^ \ frac {1} {2} U_1 }\\\\=\ Sen {ca ^ \ frac {1} {2} U_1 }_f ^ 2 \\\& = \ sen {U_1 ^ * a ^ \ frac {1} {2} c} _ f ^ 2 \\\\=\ Sen {U_1 ^ * a ^ \ frac {1} {2} Cu} _ f ^ 2 \\\&=\ Sen {U_1 ^ * a ^ \ frac {1} {2} (u_1d_1, u_2d_2 )} _ f ^ 2 \\\\\sen {U_1 ^ * a ^ \ frac {1} {2} u_1d_1 }_f ^ 2 + \ Sen {U_1 ^ * a ^ \ frac {1} {2} u_2d_2} _ f ^ 2 \ & \ geq \ Sen {U_1 ^ * a ^ \ frac {1} {2} u_1d_1} _ f ^ 2 + \ mu_k ^ 2 \ Sen {U_1 ^ * a ^ \ frac {1} {2} u_2} _ f ^ 2. \ EEA \ eeex $

 

(2 ). when $ r \ geq 1 $, make $ g (t) = t ^ r = f ^ {-1} (t) $, $ F (t) = t ^ \ frac {1} {r} $, then by (1), $ \ Bex S (F [G (A) + g (B)]) \ prec_w S (F [G (A)] + F [G (A)]) = S (A + B ). \ EEx $ by $ G $ non-negative incremental convex function, theorem 3.21, and ing theorem similar to Theorem 1.3, $ \ Bex S (G () + g (B) \ prec_ws (G (A + B )). \ EEx $ according to the fan's dominant principle, the conclusion is obtained.

[Zhan Xiang matrix theory exercise reference] exercise 4.17