inner product

bamby · #1 $Old$ November 12th, 2008, 08:54 PM

Can you help me with this. I don’t know what to do
Let R, S, T are 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.

Thanks

Opalg · #2 $Old$ November 13th, 2008, 02:19 PM

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 $\sigma\|v\|\leqslant\|Tv\|\leqslant\sigma^*\|v\|$ .