[Zhan Xiang matrix theory exercise reference] exercise 4.7

7. set $ a_0 \ In M_n $ positive, $ a_ I \ In M_n $ semi-positive, $ I = 1, \ cdots, K $, then $ \ Bex \ tr \ sum _ {j = 1} ^ k \ sex {\ sum _ {I = 0} ^ ja_ I} ^ {-2} a_j <\ tr a_0 ^ {-1 }. \ EEx $

 

 

 

Proof: note $ \ Bex \ sum _ {I = 0} ^ J a_ I = B _j, \ EEx $ then $ \ beex \ Bea \ tr \ sex {\ sum _ {I = 0} ^ J a_ I} ^ {-2} a_j & = \ tr \ sex {B _j ^ {-2} (B _j-B _ {J-1 })} \ & =\ tr (B _j ^ {-1}-B _j ^ {-2} B _ {J-1 }) \ & =\ sum _ {I = 1} ^ n s_j (B _j ^ {-1})-\ tr (B _j ^ {-2} B _ {J-1 }) \ & =\ sum _ {I = 1} ^ n s_j (B _j ^ {-1} B _ {J-1} ^ \ frac {1} {2} \ cdot B _ {J-1} ^ {-\ frac {1} {2 }}) -\ tr (B _j ^ {-2} B _ {J-1 }) \ & \ Leq \ frac {1} {2} \ sum _ {I = 1} ^ n s_ I \ sex {B _ {J-1} ^ \ frac {1} {2} B _j ^ {-2} B _ {J-1} ^ \ frac {1} {2} + B _ {J-1} ^ {-1}-\ tr (B _j ^ {-2} B _ {J-1 }) \ quad \ sex {\ mbox {inference 4.18 }\\\&=\ frac {1} {2} \ tr \ sex {B _ {J-1} ^ \ frac {1} {2} B _j ^ {-2} B _ {J-1} ^ \ frac {1} {2} + B _ {J-1} ^ {-1}-\ tr (B _j ^ {-2} B _ {J-1 }) \\&=\ frac {1} {2} \ tr (B _j ^ {-2} B _ {J-1 }) + \ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ tr (B _j ^ {-2} B _ {J-1 }) \ quad \ sex {\ tr (xy) = \ tr (Yx )} \ & =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr (B _j ^ {- 2} B _ {J-1 }) \ & =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr (B _j ^ {- 2} (B _j-A_j )) \ & =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr B _j ^ {-1} + \ frac {1} {2} \ tr (B _j ^ {-2} a_j ). \ EEA \ eeex $ hence $ \ beex \ Bea \ tr B _j ^ {-2} a_j & =\ tr B _ {J-1} ^ {-1}-\ tr B _j ^ {-1 }, \ tr \ sum _ {j = 1} ^ k \ sex {\ sum _ {I = 0} ^ ja_ I} ^ {-2} a_j & = \ sum _{ j = 1} ^ k \ tr B _j ^ {-2} a_j \ & = \ sum _ {j = 1} ^ k \ SEZ {\ tr B _ {J-1} ^ {-1}-\ tr B _j ^ {-1 }}\\\& =\ tr B _0 ^ {-1}-\ tr B _k ^ {-1 }\\& <\ tr B _0 ^ {-1 }\\&=\ tr a_0 ^ {-1 }. \ EEA \ eeex $

[Zhan Xiang matrix theory exercise reference] exercise 4.7