Verify that a Complex Vector space on $\mathbb R$ is a complex Normed Vector space in $\mathbb R$ Verify that $\mathbb C^d$ with the function $||\cdot ||_1: \mathbb C^d \to \mathbb R$ 
$||x||_1:=\sum\limits_{k=1}^d |x_k|$,    $x=(x_1,x_2,...,x_d)^T\in \mathbb C^d $
is a complex normed linear space. 
I am a bit confused with how to answer this question as I dont know if I should be should be verify the vector space axioms or not.
If I was to just show that it was a linear space, How would I do this? 
Would I have to let $y \in \mathbb C$ and then verify the normed space axioms?
Any help much appreciated.
 A: You have to verify that the map  is a norm that is that:


*

*$||x|| = 0 \iff x=0$ 

*$||x + y || \le ||x|| + ||y||$ for all $x, y \in \mathbb{C}^d$

*$|| \lambda x  ||= |\lambda| \  ||x||$ $x \in \mathbb{C}^d$ and $\lambda \in \mathbb{C}$.


It might be that you also need to show first that $\mathbb{C}^d$ is a complex vector space /linear space, in which case you have to verify that the axioms for a vectorspace hold. This works in basically the same way as it would for $\mathbb{R}^d$ in case that helps. Personally looking at the type of problem and the phrasing I would assume that you are allowed to assume as known that $\mathbb{C}^d$ is a complex vector space.
To verify that $\mathbb{C}^d$ is a $\mathbb{C}$ vectorspace one needs to check: 


*

*$(\mathbb{C}^d,+)$ is a commutative group.

*$\lambda (\mu x) = (\lambda \mu) x$ for $\lambda , \mu \in \mathbb{C}$ and $x \in \mathbb{C}^d$. 

*$1 \cdot x = x$ for  $x \in \mathbb{C}^d$. 

*$(\lambda + \mu) x = \lambda x+ \mu x $ for $\lambda , \mu \in \mathbb{C}$ and $x \in \mathbb{C}^d$

*$\lambda ( x + y ) = \lambda x+ \lambda y $ for $\lambda  \in \mathbb{C}$ and $x,y \in \mathbb{C}^d$


The operations are defined "as usual" so: $(x_1, \dots, x_d) + (y_1, \dots, y_d)=(x_1 +  y_1, \dots, x_d + y_d) $ and $\lambda (x_1, \dots, x_d)=  (\lambda x_1, \dots, \lambda x_d)$.
The argument is literally the same as for the real case. Indeed, it is the same for any field.
