# \begin{array}{ll}\min\limits_{{x,y:\,yHx=1}} x^Txyy^T\end{array}

$$x\in \mathbb{R^n}, y\in \mathbb{R^{1\times n}}$$. 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.

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

$$x^Txyy^T=||x||_2^2||y||_2^2$$, 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.