Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Possible Duplicate:
If for every $v\in V$ $\langle v,v\rangle_{1} = \langle v,v \rangle_{2}$ then $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$

Let V be a real vector space, and let $\langle x,y \rangle_1$ and $\langle x,y \rangle_2$ be two inner products defined on V.

How would you prove that if $\langle x,x \rangle_1$ = $\langle x,x \rangle_2$ $\forall x\in V$ then $\langle x,y \rangle_1$ = $\langle x,y \rangle_2$ $\forall x,y\in V$?

share|cite|improve this question

marked as duplicate by t.b., Ragib Zaman, Henning Makholm, Davide Giraudo, Jonas Meyer Nov 14 '11 at 4:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 4 down vote accepted

I think one can use polarization identity. Since given statement implies that ${||x||}_1={||x||}_2$ for all x in V. And by polarization identity we have ${\langle{x,y}\rangle}_1$ = $\frac{1}{4}{||x+y||}^{2}_{1}$- $\frac{1}{4}{||x-y||}^{2}_{1}$ = $\frac{1}{4}{||x+y||}^{2}_{2}$- $\frac{1}{4}{||x-y||}^{2}_{2}$ = ${\langle{x,y}\rangle}_2$ for all x, y in V.

share|cite|improve this answer
This is brilliant, thanks so much! – Ashbash Nov 13 '11 at 15:50

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