## Linear algebra inequality

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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.

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## 2 Answers

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.

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1
Thank you very much Robin! – Italo May 29 2010 at 14:21

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Cauchy-Schwarz in the orthogonal complement to v?

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5
Ah, Robinned again. – Charles Matthews May 29 2010 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 2010 at 13:36
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