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](https://mathoverflow.net/a/177685/32660) to [my previous question](https://mathoverflow.net/q/177683/32660) 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 $\|\Delta\|^2$.



Asymptotic property of a quadratic form
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 $A$).


