4. (G.M. krause) to $ \ bex \ lm_1 = 1, \ quad \ lm_2 = \ frac {4 + 5 \ sqrt {3} I} {13 }, \ quad \ lm_3 = \ frac {-1 + 2 \ sqrt {3} I} {13 }, \ quad v = \ sex {\ sqrt {\ frac {5} {8 }}, \ frac {1} {2 }, \ sqrt {\ frac {1} {8 }}^ T. \ eex $ then make $ \ bex A = \ diag (\ lm_1, \ lm_2, \ lm_3), \ quad U = I-2vv ^ T, \ quad B =-U ^ * AU, \ eex $ then $ U $ is the matrix, $ A, B $ is the regular matrix. verify $ \ bex \ rd (\ sigma (A), \ sigma (B) =\ sqrt {\ frac {28} {13 }}, \ quad \ sen {A-B} _ \ infty = \ sqrt {\ frac {27} {13 }}. \ eex $ therefore, for this regular matrix $ A, B $, $ \ bex \ rd (\ sigma (A), \ sigma (B)> \ sen {A-B} _ \ infty. \ eex $
Proof: Remember $ \ bex \ sigma (A) =\sed {\ al_1, \ al_2, \ al_3 }=\ sed {\ lm_1, \ lm_2, \ lm_3 }, \ quad \ sigma (B) = \ sed {\ beta_1, \ beta_2, \ beta_3 }=\ sed {-\ lm_1,-\ lm_2,-\ lm_3 }. \ eex $ then $ \ beex \ bea \ sigma =\sed {1, 2, 3} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I )} | = 2, \ sigma = \ sed {1, 3, 2} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I)} | = 2, \\\ sigma =\sed {2, 1, 3} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I )} |=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {2, 3, 1} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I) }|=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {3, 1, 2} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I )} |=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {3, 2, 1} & \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I) }|=\ sqrt {\ frac {28} {13 }}. \ eea \ eeex $ hence $ \ bex \ rd (\ sigma (A), \ sigma (B) = \ sqrt {\ frac {28} {13 }}. \ eex $ the feature value of $ (A-B) ^ * (A-B) $ is calculated by mathematical software $ \ bex \ frac {27} {13 }, \ quad \ frac {27} {13}, \ quad \ frac {4} {13 }, \ eex $ and $ \ bex \ sen {A-B} _ \ infty = \ sqrt {\ frac {27} {13 }}. \ eex $
[Zhan Xiang matrix theory exercise reference] exercise 5.4