If $A$ is real matrix of size n, and if we take the euclidian norms and the matricial norms associated, how can one show that the norm of $A $ as seen as a complex matrix is the same norm as the one of $A $ seen as a real matrix?
One inequality is trivial but I can't figure out the other one... This must be pretty easy though.....
$||A||_R^2=\sup_{||x||=1,x\in\mathbb{R}^n}x^TA^TAx$. There is $P\in O(n)$ s.t. $A^TA=P^TDP$ where $D=diag(\sigma_1^2,\cdots,\sigma_n^2)$ and $\sigma_1\geq\cdots\geq\sigma_n\geq 0$. Thus $||A||_R^2=\sup_{||x||=1,x\in\mathbb{R}^n}(Px)^TD(Px)=\sup_{||y||=1,y\in\mathbb{R}^n}y^TDy=\sup_{||y||=1,y\in\mathbb{R}^n}\sum_i\sigma_i^2 y_i^2$ (indeed, $P$ is a bijection over $\mathbb{R}^n$ s.t. $||Px||=||x||$).
In the same way, $P$ is a bijection over $\mathbb{C}^n$ and $||A||_C^2=\sup_{||z||=1,z\in\mathbb{C}^n}z^*Dz=\sup_{||z||=1,z\in\mathbb{C}^n}\sum_i\sigma_i^2 |z_i|^2\leq \sigma_1^2\sum_i |z_i|^2=\sigma_1^2$.