Q: Solution for a Frobenius norm inequality Am trying to find a real scalar $\gamma$ such that for a given pair of real rectangular matrices $X,Y$ the following holds:
$\frac{||Y||_{F}^{2}}{5} \leq ||\gamma X||_{F}^{2}\leq ||Y||_{F}^{2}$
Would it be $\gamma=\frac{\sqrt{\alpha}}{||X||_F}$ for any $\alpha$ such that 
$\frac{||Y||_{F}^{2}}{5||X||_{F}^{2}} \leq \alpha \leq \frac{||Y||_{F}^{2}}{||X||_{F}^{2}}$ ? Also let me know if there is any condition over $||X||_F,||Y||_F$ that needs to be considered while solving this problem. Thanks.

A: If $Y=0$ then either $X=0$ or $\gamma = 0$. If $X=0$ then you must have $Y=0$ (and $\gamma$ is arbitrary, of course).
Otherwise, you want $\frac{1}{5} \frac{\|Y\|^2}{\|X\|^2} \leq |\gamma|^2 \leq \frac{\|Y\|^2}{\|X\|^2}$, so any $\gamma$ that satisfies $ |\gamma| \in [\sqrt{\frac{1}{5}} \frac{\|Y\|}{\|X\|}, \frac{\|Y\|}{\|X\|}]$ will do.
