Prove Operator Norm is a Norm on linear space Prove that the operator norm defined by $$\left \| A \right \| = \left \| A \right \|_{V\rightarrow W} = \sup_{0\neq v\in V} \frac{\left \| Av \right \|_{W}}{\left \| v \right \|_{V}}$$
(Given norms $\left \| \cdot  \right \|_{V}$ and $\left \| \cdot  \right \|_{W}$ on vector spaces $V$ and $W$, and an operator $A:V\rightarrow W$)
is a norm on the linear space of operators from $V$ to $W$.
That question (Prove that the operator norm is a norm) about operator norms but in no way answers my question


First $\|A\|\ge 0$ and $\|A\|=0$ if and only if $\|A\|=0$. Surely
  $$\|A\|=\sup\frac{\|Av\|_{W}}{\|v\|_{V}}\ge0$$
  and it is equal to zero if and only if $\|Av\|_{W}=0, \forall v \Rightarrow Av=0, \forall v$ because $\|\|_{W}$ is a norm then the firs condition was proofed. 
  Now let $a\in C$ where $C$ is the field which $V$ is over then 
  $$\|aA\|=\sup\frac{\|aAv\|_{W}}{\|v\|_{V}}=\sup|a|\frac{\|Av\|_{W}}{\|v\|_{V}}=|a|\sup\frac{\|Av\|_{W}}{\|v\|_{V}}=|a|\|A\|$$
  Now let we show the last one, the triangular inequality 
  $$\|A+B\|=\sup\frac{\|(A+B)v\|_{W}}{\|v\|_{V}}\le\sup\frac{\|Bv\|_{W}+\|Av\|_{W}}{\|v\|_{V}}\le\\ \sup\frac{\|Av\|_{W}}{\|v\|_{V}}+\sup\frac{\|Bv\|_{W}}{\|v\|_{V}}=\|A\|+\|B\|$$
  I am sorry by omit the sub-script but I hope you understand that it was by simplicity. 

