MathOverflow is a question and answer site for professional mathematicians. 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

I'm wondering (hoping) if an inequality is true. Please can anyone help me?

Let $V$ be a complex vector space $dim_{\mathbb{C}}(V)=n$ with a hermitian scalar product $h$. Let $v,a, b \in V$.

Is it true that

$(h(v,v)h(a,a)-{|h(v,a)|}^{2})(h(v,v)h(b,b)-{|h(v,b)|}^{2})\geq |(h(v,v)h(a,b)-h(a,v)\overline{h(b,v)}|^{2}$?

With the overline meaning complex conjugate.

Thank you in advance.

share|cite|improve this question
up vote 4 down vote accepted

Yes. The case where $v=0$ is trivial so suppose $v\ne0$. Consider the projection map from $V$ to the hyperplane orthogonal to $v$ and let $a'$ and $b'$ be the images of $a$ and $b$ under this projection. Then your inequality reduces to $$h(a',a')h(b',b')\ge\vert h(a',b') \vert^2,$$the Cauchy-Schwarz inequality.

share|cite|improve this answer
1  
Thank you very much Robin! – Italo May 29 '10 at 14:21

Cauchy-Schwarz in the orthogonal complement to v?

share|cite|improve this answer
5  
Ah, Robinned again. – Charles Matthews May 29 '10 at 13:19
    
Charles, you had a plenty of time before Robin came: the question was asked long time ago. :) +1 to both (not to offend somebody). – Wadim Zudilin May 29 '10 at 13:36

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.