[Zhan Xiang matrix theory exercise reference] exercise 3.12

12. (Webster) set $ A = (A _ {IJ}) $ is a $ N $ level double random matrix with $ K $ positive elements. proof: $, \ cdots, an arrangement of N $ \ Sigma $ makes $ \ Bex \ sum _ {I = 1} ^ n \ frac {1} {A _ {I \ sigma (I )}} \ Leq K. \ EEx $

 

 

Proof: By Birkhoff theorem (page 35th), $ \ Bex a = \ sum \ al_kp ^ K, \ quad 0 \ Leq \ al_k \ Leq 1, \ quad \ sum \ al_k = 1, \ quad P ^ k \ mbox {replaced array }. \ EEx $ for any matrix $ B $, $ \ beex \ Bea B \ circ a & =\ sum \ al_k B \ CIRC P ^ K, \\\ sum _ {I = 1} ^ n B _ {IJ} A _ {IJ} & =\ sum \ al_k \ sum _ {I, j = 1} ^ n B _ {IJ} P ^ K _ {IJ} \ & \ geq \ min _ {P \ In \ pi_n} \ sum _ {I, j = 1} ^ n B _ {IJ} p _ {IJ} \ quad \ sex {\ pi_n \ mbox {A set composed of all replacement arrays }\\\&= \ sum _ {I = 1} ^ n B _ {I \ sigma (I )} \ quad \ sex {\ mbox {An arrangement related to} B \ mbox {} \ Sigma} exists }. \ EEA \ eeex $ set $ B $ to $ \ bee \ label {3_12_ B} B _ {IJ }=\ sedd {\ BA {ll} \ cfrac {1} {A _ {IJ }}, & A _ {IJ} \ NEQ 0, \ k + 1, & A _ {IJ} = 0. \ EA} \ EEE $ then $ \ bee \ label {3_12_k} k = \ sum _ {I, j = 1} ^ n B _ {IJ} A _ {IJ} \ geq \ sum _ {I = 1} ^ n B _ {I \ sigma (I )}. \ EEE $ for each $ B _ {I \ sigma (I)} \ geq 0 $, \ eqref {3_12_k} knows that $ B _ {I \ sigma (I)} $ cannot be $ k + 1 $, and \ eqref {3_12_ B }, $ \ Bex B _ {I \ sigma (I) }=\ frac {1} {A _ {I \ sigma (I )}}. \ EEx $ so, \ eqref {3_12_k} becomes $ \ Bex k \ geq \ sum _ {I = 1} ^ n \ frac {1} {A _ {I \ sigma (I )}}. \ EEx $

[Zhan Xiang matrix theory exercise reference] exercise 3.12