most recent 30 from http://mathoverflow.net 2013-05-23T13:32:40Z http://mathoverflow.net/feeds/question/26350 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26350/linear-algebra-inequality Italo 2010-05-29T11:45:02Z 2010-06-22T21:40:18Z

<p>I'm wondering (hoping) if an inequality is true. Please can anyone help me?</p> <p>Let $V$ be a complex vector space $dim_{\mathbb{C}}(V)=n$ with a hermitian scalar product $h$. Let $v,a, b \in V$.</p> <p>Is it true that</p> <p>$(h(v,v)h(a,a)-{|h(v,a)|}^{2})(h(v,v)h(b,b)-{|h(v,b)|}^{2})\geq |(h(v,v)h(a,b)-h(a,v)\overline{h(b,v)}|^{2}$?</p> <p>With the overline meaning complex conjugate.</p> <p>Thank you in advance.</p>

http://mathoverflow.net/questions/26350/linear-algebra-inequality/26353#26353 Robin Chapman 2010-05-29T13:17:00Z 2010-05-29T13:17:00Z

<p>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.</p>

http://mathoverflow.net/questions/26350/linear-algebra-inequality/26354#26354 Charles Matthews 2010-05-29T13:18:39Z 2010-05-29T13:18:39Z

<p>Cauchy-Schwarz in the orthogonal complement to v?</p>