How to prove the following equality with matrix norm? $\|(H+\phi I)^{-1}(H-\varphi I)\|_2^2=\max_{\|x\|_2=1}\frac{\|(H-\varphi I)x\|_2^2}{\|(H+\phi I)x\|_2^2}$,
where both H is a nonsymmetric matrix and $\phi$ and $\varphi$ are two positive parameters. I could not derive it from the definition. 
Is it also valid for $\|(H+\phi I)^{-1}(V-\varphi I)\|_2^2=\max_{\|x\|_2=1}\frac{\|(V-\varphi I)x\|_2^2}{\|(H+\phi I)x\|_2^2}$, where $H$ and $V$ are not commute (HV=VH does not hold)?
 A: You have 
\begin{align}
\sup\{\|(H+\phi I)^{-1}(H-\varphi I)x\|:\ \|x\|=1\}
&=\sup\left\{\frac{\|(H-\varphi I)(H+\phi I)^{-1}x\|}{\|x\|}:\  x\ne0\right\}\\ \ \\
&=\sup\left\{\frac{\|(H-\varphi I)x\|}{\|(H+\phi I)x\|}:\  x\ne0\right\}\\ \ \\
&=\sup\left\{\frac{\|(H-\varphi I)x\|}{\|(H+\phi I)x\|}:\  \|x|=1\right\},\\ \ \\
\end{align}
where


*

*the first equality is due to $H+\phi I$ and $H-\varphi I$ commuting;

*the second equality is due to $H-\phi I$ invertible, and so $(H-\phi I)x$ runs over all vectors when $x$ runs over all vectors;

*the third equality can be achieved by dividing numerator and denominator by $x$. 
You cannot replace one of the $H$ for an arbitrary $V$. For instance take $\phi=\varphi=0$, 
$$
H=\begin{bmatrix}1&0\\0&2 \end{bmatrix} ,\ \ \ \ V=\begin{bmatrix} 
2&2\\0&0\end{bmatrix}.
$$
Then 
$$
\|H^{-1}V\|=\left\|\begin{bmatrix} 2&2\\0&0\end{bmatrix}\right\|
=\left\|\begin{bmatrix} 2&2\\0&0\end{bmatrix}\begin{bmatrix} 2&0\\2&0\end{bmatrix}\right\|^{1/2}=\sqrt8=2\sqrt2.
$$
And, with $x_1^2+x_2^2=1$, $t=x_1$,
$$
\frac{\|Vx\|}{\|Hx\|}=\frac{2|x_1+x_2|}{\sqrt{x_1^2+2x_2^2}}
=\frac{2|x_1+\sqrt{1-x_1^2}|}{\sqrt{1+x_2^2}}
=\frac{2(t+\sqrt{1-t^2})}{2-t^2}\leq\sqrt6<\sqrt8=\|H^{-1}V\|.
$$
