Let $M\in M_n(\mathbb C)$ be a $n\times n$ matrix over the complex field. It can be written uniquely as $M=H+A$, where $H=H^*$ denotes its Hermitian part and $A=-A^*$ its anti-Hermitian part.

Its spectral norm is defined as \begin{equation} ||M||^2=\underset{x}{\text{max}}\,\frac{(Mx,Mx)}{(x,x)}\;, \end{equation} where the maximum is taken over all nonzero vectors $x\in \mathbb C^n$ and $(y,x)$ denotes the standard inner product in $\mathbb C^n$, i.e. $(y,x)\equiv \sum_{i=1}^n \bar y_i x_i$. The spectral norm is well-known to be equal to the square root of maximum eigenvalue of $M^* M$.

Consider matrices $M_U=U^*HU+A$, where $U$ is an $n\times n$ unitary matrices, i.e. they all have the same fixed anti-Hermitian part and the same fixed Hermitian part's spectrum.

What is the maximum possible value which $||M_U||$ can attain upon varying $U$?

By the triangle inequality, a trivial bound is $||H||+||A||$, but this is not in general achievable. Instead, I am looking for an expression for $\text{max}_U\,||M_U||$, for general $H$ and $A$ (so as to avoid going through the maximization over all possible unitary matrices each time).

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    $\begingroup$ Can't you just choose $U$ to "rotate" things such that an eigenvector of $A$ corresponding to its largest (in modulus) eigenvalue also becomes an eignvector of$ U^HU^*$ corresponding to its largest (in modulus) eigenvalue? $\endgroup$ – Yemon Choi Oct 1 '17 at 16:06
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    $\begingroup$ In which case I think you get $(\Vert H\Vert^2+\Vert A\Vert^2)^{1/2}$ $\endgroup$ – Yemon Choi Oct 1 '17 at 16:07
  • $\begingroup$ I am not familiar with the notation $U^H$. What does it mean? $\endgroup$ – francesco999 Oct 1 '17 at 19:55
  • $\begingroup$ Also, it would be great if you could spell out maybe in more detail how you get to $(||H||^2+||A||^2)^{1/2}$ $\endgroup$ – francesco999 Oct 1 '17 at 19:57
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    $\begingroup$ I'm not sure one can give a very general answer, it'll depend on what $A,H$ are. If both matrices have one eigenvalue that is much larger than all the others, then what Yemon suggests feels about right. If they have eigenvalues of comparable (or equal) size, then one has more options and can do better. $\endgroup$ – Christian Remling Oct 1 '17 at 21:41

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