$x\in \mathbb{R^n}, y\in \mathbb{R^{1\times n}}$. Given $H\in \mathbb{R^{n\times n}}$ \begin{array}{ll}\min\limits_{{x,y:\,yHx=1}} x^Txyy^T\end{array}

My attepmt:

SVD: $H=U\Sigma V^*$. Let $x=\frac{1}{\sigma_1}v_1$ and $y=u_1^*$, where $u_1$ and $v_1$ are the first columns of $U$ and $V$ respectively, and $\sigma_1$ is the largest singular value of $H$. Then $RHT=1$. \begin{array}{ll}\min\limits_{{x,y:\,yHx=1}} x^Txyy^T=\frac{1}{\sigma_1^2}.\end{array}

I tried to rewrite the problem in terms of traces: Let $xy=Z$, \begin{array}{ll}\min\limits_{{x,y:\,yHx=1}} x^Txyy^T=\min\limits_{{Z:\,tr(HZ)=1,\\\,\,\,\,Z \,\text{is rank 1}}} tr(ZZ^T)\end{array} I believe we need to use the fact that $Z$ has only 1 nonzero eigenvalue.

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    $\begingroup$ Are you sure it's max and not min? $\endgroup$ – eyeballfrog Sep 4 '19 at 5:45
  • $\begingroup$ @eyeballfrog, yes indeed it was min, I have changed it already. Thanks for the observation $\endgroup$ – Lee Sep 4 '19 at 6:44
  • $\begingroup$ @eyeballfrog I have copied the equation from latex to modify, forgot to change max to min $\endgroup$ – Lee Sep 4 '19 at 6:46

My friends helped me to answer this question. I will post it here, in case someone will need it:

$x^Txyy^T=||x||_2^2||y||_2^2$, where $||\cdot||_2$ is second norm. By Cauchy Schwarz inequality: $||y||_2\cdot||H||_2\cdot||x||_2\geq yHx=1$, Note: here $||H||_2$ is induced second norm.

Then $||y||_2\cdot||H||_2\cdot||x||_2=\sigma_1\cdot||y||_2\cdot||x||_2\geq1,$ thus $||x||_2^2||y||_2^2\geq\frac{1}{\sigma_1^2}$. And in my question I have already showed that the equality is attainable.


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