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suppose $x=\Delta$, $y=M \Phi \Delta$, where $\Delta\in N\times 1$, $M^T=M \in N \times N$ and $\Phi^T=\Phi \in N \times N$. Define $Z=xy^T+yx^T$. It is known from the answer to my previous question that $Z$ has two eigenvalues, one is positive and the other is negative and they are given by

$y^Tx+\sqrt{x^Txy^Ty}$ and $y^Tx-\sqrt{x^Txy^Ty}$.

My question is if we have the following result:

$y^Tx+\sqrt{x^Txy^Ty}=\mathcal{O}(\|\Delta\|^2)$, as $\|\Delta\|\rightarrow \infty$.

Based on my simulations, if increases the magnitude of the elements of $\Delta$, $y^Tx+\sqrt{x^Txy^Ty}$ grows linearly w.r.t. the increases of $\|\Delta\|^2$.

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Not sure if I understand your question correctly.

Denote $A=M\Phi$. Then for $x$ nonzero, $$\frac{y^Tx+\sqrt{x^Txy^Ty}}{\|x\|^2}=\frac{x^TA^Tx}{x^Tx}+\sqrt{\frac{x^TA^TAx}{x^Tx}}.$$

As $A$ is a finite dimensional matrix, $\frac{x^TA^Tx}{x^Tx}+\sqrt{\frac{x^TA^TAx}{x^Tx}}$ is bounded above by a constant, (the numerical radius of $A$ plus the largest singular value of $A$).

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$E=y^Tx+\sqrt{x^Txy^Ty}=||Z||_2\leq 2||x||_2||y||_2\leq 2||M||_2||\phi||_2{||\Delta||_2}^2$.

@ Michael Fan Zhang , The Lin's answer is correct. Why do you give no points to him ?

EDIT: $E=\Delta^T\phi M\Delta+||\Delta||||M\phi\Delta||$. Assume that , for every $u\not= 0$, $u^T\phi Mu>0$. In particular, $\phi M$ and $M\phi$ are invertible; moreover the real part of any eigenvalue of $\phi M$ is necessarily positive. Then there is $\alpha >0$ s.t. $E\geq \alpha ||\Delta||^2$.

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  • $\begingroup$ Hi Loup, thanks for your answer and reminder. BTW, any lower bound with quadratic growth for the expression? $\endgroup$ – Michael Fan Zhang Sep 17 '14 at 5:24

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