4. set $ A = (A _ {IJ}) \ In M_n $, then $ \ Bex \ sex {| A _ {11} |, \ cdots, | A _ {NN} |} \ prec_ws (). \ EEx $
Proof: in general, we use the fan's dominant principle in the following scenarios: $ \ Bex S (a) \ prec S (B) \ LRA \ mbox {to any undo norm} \ Sen {\ cdot}, \ Sen {A} \ Leq \ Sen {B }. \ EEx $, but here we need to use reverse push. it can be seen that fan is great. note $ \ Bex \ Omega = e ^ {\ frac {2 \ Pi I} {n}, \ quad I = \ SQRT {-1 }, \ quad u = \ diag (1, \ Omega, \ Omega ^ 2, \ cdots, \ Omega ^ {n-1}), \ EEx $ then $ U $ is a array, and $ \ bee \ label {4_4_diag} \ diag (A _ {11}, \ cdots, A _ {NN }) = \ frac {1} {n} \ sum _ {k = 0} ^ {n-1} U ^ Kau ^ {* k }. \ EEE $ \ eqref {4_4_diag} can be obtained by comparing the elements of the matrix. in fact, $ (I, j) of the right-end matrix of \ eqref {4_4_diag) $ element: $ \ beex \ Bea \ frac {1} {n} \ sum _ {k = 0} ^ {n-1} \ Omega ^ {ik} A _ {IJ} \ bar \ Omega ^ {JK} & =\ frac {1} {n} \ sum _ {k = 0} ^ n \ Omega ^ {ik} \ bar \ Omega ^ {JK} A _ {IJ} \\\&=\ sedd {\ BA {ll} A _ {II }, & I = J \\\ cfrac {1} {n} \ DPS {\ sum _ {k = 0} ^ {n-1} \ Omega ^ {(I-j) k} A _ {IJ }}=\ cfrac {1} {n} \ cdot \ cfrac {1-\ Omega ^ {(I-j) N }}{ 1-\ Omega ^ {I-j} A _ {IJ} = 0, & I \ NEQ J \ EA }\\& = A _ {IJ} \ Delta _ {IJ }. \ EEA \ eeex $ by \ eqref {4_4_diag}, you can call this operation to know any Unio constant norm $ \ Sen {\ cdot} $, $ \ Bex \ Sen {\ diag (A _ {11}, \ cdots, A _ {NN })} \ Leq \ frac {1} {n} \ sum _ {k = 0} ^ {n-1} \ Sen {u ^ Kau ^ {* k }}=\ Sen {}. \ EEx $ according to the fan principle, $ \ Bex \ sex {| A _ {11} |, \ cdots, | A _ {NN} |} \ prec_ws (). \ EEx $
[Zhan Xiang matrix theory exercise reference] exercise 4.4