TITLE: unitization of a unital $C^*$-algebra
QUESTION [9 upvotes]: I have a little question about unitization of a $C^*$-algebra.
If $A$ is a non-unital $C^*$-algebra, set $A_1=A\oplus\mathbb{C}$ as vector spaces and define a multiplikation, involution and a norm in a usual way,
$$(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu)$$ 
$$(a,\lambda)^*=(a^*,\overline{\lambda})$$ and $$\|(a,\lambda)\|=\|L_{(a,\lambda)}\|,$$ with $L_{(a,\lambda)}:A\to A,\; b\mapsto ab+\lambda b$ is the multiplication operator. You will obtain a $C^*$-algebra $A_1$ with unit $(0,1)$ and $A$ can be identified with $\{(a,0);a\in A\}$, $A$ is a closed ideal in $A_1$. My question is, what exanctly changes if $A$ is unital and you consider it's unitization $A_1$?
If $A$ is unital, its unitization $A_1$ is again $A\oplus\mathbb{C} $ as vector spaces and the involution is the same. But the norm seems to be different, and why? What else is different in this case?
Edit: Ok, my problem is tho following:
Frequently, if the unitization of $A$ will be introduced, it is assumed that A is non-unital, for example here: http://ncatlab.org/nlab/show/unitization+of+a+C-star-algebra . But if I want to show that $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|$ is a norm, I need that A is non-unital to prove: if $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|=0, \Rightarrow (a,\lambda)=(0,0)$.
Therefore my question is, what is the correct $C^*$-morm on $A_1$, if $A$ is unital? For example I read $\|(a,\lambda)\|=max\{\|a\|,|\lambda|\}$ should be the $C^*$-Norm on $A_1$ if A is unital. But I dont see how it could be, because $\|(a,\lambda)\|=max\{\|a\|,|\lambda|\}$ is completely different as $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|$. 
To put my question in a nutshell: What exactly is the $C^*$-norm on $A_1$ if $A$ is unital and non-unital?
An other thing is the multiplication in the non-unital case and the unital case. If $A$ is a Banach algebra, the multiplication on $A_1$ should be $(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu)$, but I read that if $A$ is unital, than the multiplication is $(a,\lambda)\cdot (b,\mu)=(ab,\lambda\mu)$. But I don't know why..

REPLY [2 votes]: Same issue arises in plain Banach algebras (except then you can be a little sloppier defining the norm on the unitization). If $A$ is a Banach algebra with identity and you define $A_1$ as above then $(0,1)$ is the identity in $A_1$. The previous identity $e$, which looks like $(e,0)$ in $A_1$, is no longer an identity, because for example $(e,0)(0,1)=(e,0)$. 
You ask why the norm seems to be different. Not sure what you mean - if you mean the norm on $A_1$ is different from the norm on $A$, yes it is. It's a different space. If you mean the norm works out differently in this case from how it works out for non-unital $A$ I don't see what difference you're referring to.