how to interpret this norm

Question

Could any one tell me what "x" is there when he has defined $\|x\|$, just after he says $M_n$ has the operator norm

thank you for helping.

Answer 1

The concept of operator norms becomes more clear when we consider normed vector spaces $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$. The space of linear maps from $X$ to $Y$, $L(X,Y)$ is equipped with the operator norm $$\| f \|_L = \sup_{x\in X} \frac{\|f(x)\|_Y}{\|x\|_X} = \sup_{\|x\|_X = 1} \|f(x)\|_Y$$ Now the special case is $(X,\|\cdot\|_Y) = (Y,\|\cdot\|_Y) = (\mathbb C^n, |\cdot|)$ and we have $L(\mathbb C^n, \mathbb C^n) \simeq \mathbb C^{n\times n}$ the ring of complex $n\times n$ matrices with norm $$\| A \|_L = \sup_{|z|=1} |Az|$$

Answer 2

Here $x\in \Bbb M_n$. So operator norm of $M_n$ is calculated.

Answer 3

$x:\mathbb{C}^n\to \mathbb{C}^n$ is a complex linear map. The norm $\|x\|$ can be written as:

$$\|x\|=\sup_{|\xi|=1} |x\xi|.$$ That is, the norm of the given linear is the maximum on the unit sphere (identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$). (We can say maximum because of continuity and compactness). In other words, $\|x\|$ it is the maximum of the modules of its eigenvalues.

Answer 4

The author wants to define an operator norm on $\mathbb M_n$, that is a map $\mathbb M_n\to \mathbb R$. As often when defining a function $f\colon A\to B$ one writes down $f(x)=\text{some expression}$, except that in the case of a norm one conventionally writes $\|x\|$ instead of $f(x)$. Thus the equation defines the operator norm considered by specifying its value on an arbitrary element $x\in\mathbb M_n$.