Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the best way to prove the polarization identity in the following form: for any $u, v \in V$ $$4 \langle T(u), v \rangle = \langle T(u+v), u+v \rangle - \langle T(u-v), u-v \rangle + i\langle T(u+iv), u+iv \rangle - i\langle T(u-iv), u-iv \rangle$$

Here $T$ is a linear operator.

Context

Polarization identity is a useful result in Hilbert spaces. It expresses the inner product in terms of norm:
$$4 \langle u, v \rangle = \langle u+v, u+v \rangle - \langle u-v, u-v \rangle + i\langle u+iv, u+iv \rangle - i\langle u-iv, u-iv \rangle$$ where all terms on the right are just norms squared.

More generally, it gives a way to estimate a bilinear form $B(u,v) = \langle T(u), v \rangle $ in terms of quadratic form $Q(u)=B(u,u)$. This is the generalized form with $T$ above.

share|cite|improve this question
    
What is the property of the $\langle . , . \rangle$ operator? $\langle u, v + w \rangle = ...$ (Hint: in the real case, we speak about bilinear forms). – Djaian Nov 4 '10 at 13:59

Expand RHS, using distributive property and the property that $\langle u,cv\rangle=\bar{c}\langle u,v\rangle$. Simplify and cancel many terms to get the LHS.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.