1. Set $ A \ In M_n $. Prove that if $ AA ^ * = a ^ 2 $, $ A ^ * = A $.
Proof: By Schur ry triangle theorem, $ U $ exists, making $ \ Bex a = u ^ * Bu, \ EEx $ where $ B = (B _ {IJ}) $ is the upper triangle array. so $ \ Bex U ^ * BB ^ * u = AA ^ * = a ^ 2 = u ^ * B ^ 2u \ rA BB ^ * = B ^ 2. \ EEx $ compare the diagonal yuan between two ends $ \ Bex | B _ {II} | ^ 2 + \ cdots + | B _ {In} | ^ 2 = B _{ II} ^ 2, \ Quad 1 \ Leq I \ Leq n. \ EEx $ while $ B _ {II} ^ 2 $ is a non-negative real number, $ \ Bex B _ {II} ^ 2 = | B _ {II} | ^ 2 + \ cdots + | B _ {In} | ^ 2 \ geq | B _{ II} | ^ 2 = | B _ {II} ^ 2 | = B _ {II} ^ 2. \ EEx $ hence $ \ Bex B _ {IJ} = 0, \ quad I <J, \ quad B _ {II }=\ PM \ SQRT {B _ {II} ^ 2} \ In \ BBR. \ EEx $ therefore, $ \ Bex a = u ^ * \ diag (B _ {11}, \ cdots, B _ {NN}) U, \ quad B _ {II} \ In \ BBR. \ EEx $ this indicates $ A ^ * = A $.
[Zhan Xiang matrix theory exercise reference] exercise 3.1