Let $M_n(\mathbb{C})$ denote the algebra of $n \times n$ complex matrices, and $H_n(\mathbb{C})$ denote the linear space of $n \times n$ Hermitians. Both spaces are endowed with the usual operator norm. Assume that
$$ \Phi \colon H_n(\mathbb{C}) \rightarrow H_n(\mathbb{C})$$
is a (real) linear map of norm $1.$
Then $\Phi$ has a natural linear extension to $M_n(\mathbb{C})$ by
$$\Phi(A) := \Phi\left({A+A^* \over 2}\right) + i\Phi\left({A-A^* \over 2i}\right).$$
Could you give me an example with $\|\Phi\| > 1 ?$ Or an explanation why this might happen?
Best Answer
I think I have an example showing the norm can be greater than $1$. For Hermitian $H=\begin{bmatrix}a&b\\\overline{b}&d\end{bmatrix}\in H_2(\mathbb C)$, define $\Phi(H)=\begin{bmatrix}0&\dfrac{a+di}{\sqrt2}\\\dfrac{a-di}{\sqrt2}&0\end{bmatrix}.$ We have $\|\Phi(H)\|=\frac1{\sqrt2}\sqrt{a^2+d^2}\leq\max\{|a|,|d|\}\leq\|H\|,$ with equality holding in case $|a|=|d|$ and $b=0$.
Consider the extension applied to $A=\begin{bmatrix}1&0\\0&i\end{bmatrix}$. We have $\|A\|=1$ while $\tilde{\Phi}(A)=\begin{bmatrix}0&0\\\sqrt2&0\end{bmatrix},$ so $\|\tilde{\Phi}(A)\|=\sqrt2$.
Motivation: To have the norm of the extension increase, it makes sense to look for cases where the real and imaginary parts of $A$ are "nonoverlapping," so that $A+A^*$ and $A-A^*$ can be as large as possible without increasing the norm of $A$. Then if we can have $\Phi$ map each of those parts in such a way that they do "overlap," an increase in norm can happen. This was done by taking distinct diagonal entries and placing them in the same off-diagonal positions. The off-diagonal was needed to allow the diagonal entries from the Hermitian matrices to be sent to real and imaginary parts, keeping the norm of $\Phi$ from being too large on $H_2(\mathbb C)$.