[Zhan Xiang matrix theory exercise reference] exercise 4.3

3. $ g \ In M_n $ is called a rank $ K $ partial offset matrix. If $ \ Bex S_1 (G) = \ cdots = S_k (G) = 1, \ quad S _ {k + 1} (G) = \ cdots = s_n (G) = 0. \ EEx $ proof pair $ x \ In M_n $, $ \ Bex \ sum _ {j = 1} ^ K s_j (X) = \ MAX \ sed {| \ tr (xg) |; G \ mbox {is a rank} k \ mbox {partial offset matrix,} g \ In M_n }. \ EEx $ use this expression to prove theorem 4.9.

 

 

 

Proof: (1 ). set $ x $ to Singular Value Decomposition $ \ Bex x = U \ diag (S_1, \ cdots, s_n) V, \ EEx $ where $ u, v $ are all arrays. $ \ Bex G = V ^ * \ diag (\ underbrace {1, \ cdots, 1 }_{ k \ mbox {}}, 0, \ cdots, 0) U ^ *, \ EEx $ \ beex \ Bea | \ tr (xg) | & = | \ tr (U \ diag (S_1, \ cdots, S_k, 0, \ cdots, 0) U ^ * | \\&=| \ tr (\ diag (S_1, \ cdots, S_k, 0, \ cdots, 0 )) | \\& =\ sum _ {j = 1} ^ K s_j (X ). \ EEA \ eeex $(2 ). for any rank $ K $ partial offset matrix $ G $, $ \ beex \ Bea | \ tr (xg) | & \ Leq \ sum _ {j = 1} ^ n s_j (xg) \ quad \ sex {\ mbox {inference 4.11 }\\\& \ Leq \ sum _ {j = 1} ^ n s_j (x) S_1 (g) \ quad \ sex {\ mbox {Theorem 4.3 }\\\&=\ sum _ {j = 1} ^ n s_j (X ). \ EEA \ eeex $(3 ). the proof theorem 4.9 is as follows. for $1 \ Leq k \ Leq N $, $ \ beex \ Bea \ sum _ {I = 1} ^ K s_ I (A + B) & =\ sum _ {I = 1} ^ k \ MAX \ sed {| \ tr (a + B) G) |; g \ mbox {is a rank} k \ mbox {partial offset matrix ,} g \ In M_n }\\\& \ Leq \ sum _ {I = 1} ^ k \ MAX \ sed {|\ tr (AG) |; g \ mbox {is a rank} k \ mbox {partial offset matrix ,} g \ In M_n }\\\& \ quad + \ sum _ {I = 1} ^ k \ MAX \ sed {|\ tr (BG) |; g \ mbox {is a rank} k \ mbox {partial offset matrix,} g \ In M_n }\\\&=\ sum _ {I = 1} ^ K s_ I () + \ sum _ {I = 1} ^ K s_ I (B) \\\&=\ sum _ {I = 1} ^ K [s_ I () + s_ I (B)]. \ EEA \ eeex $

[Zhan Xiang matrix theory exercise reference] exercise 4.3