How to calculate $\left\|\left(\begin{array}{}0&F\\F^*&0\end{array}\right)\right\|$ in $M_2(A)$?

Question

My approach: let $v$ be a partial isometry, outside $A$, such that $v^*va=av^*v=a$ and $vv^*a=avv^*=0$ and $vv=0$ for every $a\in A$. Then by defining $\varphi\left(\begin{array}{}a&b\\c&d\end{array}\right)=a+bv+v^*c+v^*dv$ we see $M_2(A)$ is isomorphic to $C^*(A,v)$. Thus I can calculate $\|Fv+v^*F^*\|$ instead.

Firstly, $\|Fv+v^*F^*\|\geq \|vFv+vv^*F^*\|=\|F^*\|=\|F\|$

Secondly, $\|Fv+v^*F^*\|^{2^{n+1}}=\|(FF^*)^{2^n}+v^*(FF^*)^{2^n}v\|\leq 2\cdot \|F\|^{2^{n+1}}$

Thus I conclude $\|Fv+v^*F^*\|=\|F\|$.

What is the standard way...?

Answer 1

The standard way is \begin{align} \left\|\begin{bmatrix} 0&F\\ F^*&0\end{bmatrix}\right\|^2 &= \left\|\begin{bmatrix} 0&F\\ F^*&0\end{bmatrix}\begin{bmatrix} 0&F\\ F^*&0\end{bmatrix}\right\| = \left\|\begin{bmatrix} FF^*&0\\ 0&F^*F\end{bmatrix}\right\|\\[0.3cm] &=\max\{\|FF^*\|,\|F^*F\|\}=\|F\|^2. \end{align}