Thread: inner product
View Single Post
  #2  
Old November 13th, 2008, 02:19 PM
Opalg's Avatar
Opalg Opalg is offline
MHF Contributor

 
Join Date: Aug 2007
Location: Leeds, UK
Posts: 2,063
Country:
Thanks: 126
Thanked 1,258 Times in 1,046 Posts
Opalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant futureOpalg has a brilliant future
Default

Quote:
Originally Posted by bamby View Post
Let R, S, T be linear operators, where V is a complex inner product space.
(i) Suppose that S is an isometry and R is a positive operator such that T=SR. Prove that R=square root of (T*T)
(ii) Let σ denote the smallest singular value of T, and let σ*denote the largest singular value of T. Prove that σ<=|| T(v)/||v|| ||<= σ* for every nonzero v in V.
(i) If S is an isometry then S*S = I. Therefore T^*T = RS^*SR = R^2. But a positive operator has a unique positive square root, so R=(T^*T)^{1/2}.

(ii) \|Tv\|^2 = \langle T^*Tv,v\rangle = \langle R^2v,v\rangle. But \sigma^2 I\leqslant R^2\leqslant \sigma^{*2}I. So \sigma^2\langle v,v\rangle \leqslant\langle R^2v,v\rangle\leqslant \sigma^{*2}\langle v,v\rangle, from which \sigma\|v\|\leqslant\|Tv\|\leqslant\sigma^*\|v\|.

Last edited by Opalg; November 14th, 2008 at 02:17 AM. Reason: corrected typo
Reply With Quote
The following users thank Opalg for this useful post:
Donate to MHF