[Zhan Xiang matrix theory exercise reference] exercise 3.11

11. (Ky Fan) for $ A \ In M_n $, remember $ \ re a = (a + A ^ *)/2 $. proof: $ \ Bex \ re \ LM (a) \ prec \ LM (\ re a), \ EEx $ where $ \ LM () $ represents the vector produced by the feature values of $ A $, and $ \ re \ LM (a) $ returns the real-part vector of the feature values of $ A $.

 

 

Proof:

 

(1 ). first, verify the Hermite array $ h $. If its feature value is $ \ Bex \ lm_1 \ geq \ cdots \ geq \ lm_n, \ EEx $ \ Bex \ sum _ {I = 1} ^ k \ lm_ I = \ Max _ {\ Sen {X_ I} = 1 \ atop I = 1, \ cdots, k} \ sum _ {I = 1} ^ k \ SEF {hx_ I, X_ I }. \ EEx $ in fact, the exists from $ h $ for Hermite matrix, making $ \ Bex U ^ * Hu = \ diag (\ lm_1, \ cdots, \ lm_n ). \ EEx $ note $ u = (U_1, \ cdots, u_n) $, then $ \ Bex \ sum _ {I = 1} ^ k \ SEF {hu_ I, u_ I }=\ sum _ {I = 1} ^ k \ lm_ I. \ EEx $ on the other hand, for any vector group $ x_1, \ cdots, X_k $ suitable for $ \ Sen {X_ I} = 1 $, you can set $ \ Bex X_ I = \ sum _ {j = 1} ^ N A _ {IJ} u_j, \ quad \ sum _ {j = 1} ^ n | A _ {IJ} | ^ 2 = 1, \ EEx $ and $ \ beex \ Bea ax_ I & =\ sum _ {j = 1} ^ n a _ {IJ} \ lm_ju_j, \\\ SEF {ax_ I, x_ I} & =\ SEF {\ sum _ {j = 1} ^ N A _ {IJ} \ lm_ju_j, \ sum _ {L = 1} ^ n a _ {il} u_l} \ & = \ sum _ {J, L = 1} ^ n a _ {IJ} \ bar a _ {il} \ lm_j \ SEF {u_j, u_l }\\\=\ sum _ {j = 1} ^ n a _ {IJ} \ bar a _ {IJ} \ lm_j \\\=\ sum _ {J = 1} ^ n | A _ {IJ} | ^ 2 \ lm_j \\\\=\ sum _ {j = 1} ^ n | A _ {IJ} | ^ 2 \ lm_k + \ sum _ {j = 1} ^ n | A _ {IJ} | ^ 2 (\ lm_j-\ lm_k) \\& =\ lm_k + \ sum _ {j = 1} ^ k | A _ {IJ} | ^ 2 (\ lm_j-\ lm_k) + \ sum _ {J = k + 1} ^ n | A _ {IJ} | ^ 2 (\ lm_j-\ lm_k) \\& \ Leq \ lm_k + \ sum _ {j = 1} ^ k | A _ {IJ} | ^ 2 (\ lm_j-\ lm_k ), \\\ sum _ {I = 1} ^ k \ SEF {ax_ I, x_ I} & \ Leq k \ lm_k + \ sum _ {I = 1} ^ k \ sum _ {j = 1} ^ k | A _ {IJ} | ^ 2 (\ lm_j-\ lm_k) \\& \ Leq k \ lm_k + \ sum _ {j = 1} ^ K (\ lm_j-\ lm_k) \ sum _ {I = 1} ^ k | A _ {IJ} | ^ 2 \ & \ Leq k \ lm_k + \ sum _ {j = 1} ^ K (\ lm_j-\ lm_k) \\& =\ sum _ {j = 1} ^ k \ lm_j. \ EEA \ eeex $

 

(2 ). repeat the question. by $ A \ In M_n $ and Schur ry triangle theorem, $ makes $ \ Bex V ^ * AV = \ sex {\ BA {CCC} \ lm_1 () & * \\& \ ddots \\\& \ lm_n (a) \ EA }. \ EEx $ note $ v = (v_1, \ cdots, v_n) $, then $ \ Bex \ SEF {av_ I, V_ I }=\ lm_ I (). \ EEx $ is equivalent to $1 \ Leq k \ Leq N $, $ \ beex \ Bea \ sum _ {I = 1} ^ k \ re \ lm_ I () & =\ sum _ {I = 1} ^ k \ frac {\ lm_ I (A) + \ overline {\ lm_ I ()}} {2 }\\\=\ sum _ {I = 1} ^ k \ frac {1} {2} \ SEF {av_ I, v_ I} + \ frac {1} {2} \ overline {\ SEF {av_ I, v_ I }\\\=\ sum _ {I = 1} ^ k \ frac {1} {2} \ SEF {av_ I, v_ I} + \ frac {1} {2} \ SEF {A ^ * V_ I, V_ I} \ quad \ sex {\ overline {\ SEF {av_ I, v_ I }=\ SEF {V_ I, av_ I }=\ SEF {A ^ * V_ I, v_ I }\\\=\ sum _ {I = 1} ^ k \ SEF {\ frac {A + A ^ *} {2} V_ I, v_ I }\\&=\ sum _ {I = 1} ^ k \ SEF {\ re a V_ I, v_ I }\\& \ Leq \ sum _ {I = 1} ^ k \ lm_ I (\ re a) \ quad \ sex {\ mbox {by (1 )}}. \ EEA \ eeex $

[Zhan Xiang matrix theory exercise reference] exercise 3.11