unitization of a unital $C^*$-algebra

Question

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 multiplication, 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 $\{(a,0)\mid a\in A\}$, $A$ is a closed ideal in $A_1$. My question is, what exactly changes if $A$ is unital and you consider its 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 the following:

Frequently, if the unitization of $A$ will be introduced, it is assumed that $A$ is non-unital, for example here. 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 $C^*$-norm 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 don't 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, then the multiplication is $(a,\lambda)\cdot (b,\mu)=(ab,\lambda\mu)$. But I don't know why.

Answer 1

The issue is not about the norm. If $A$ is already unital, then the unitization construction as you described it is not well-defined, in the following sense. Since $1\in A$, you can consider the element $(-1,1)$, which of course should be zero. This forces you to construct the unitization as a quotient $(A\oplus\mathbb C)/J$, where $J$ is the ideal $$ J=\{(-t,t):\ t\in \mathbb C\}. $$ And then you can easily show that $(A\oplus\mathbb C)/J\simeq A$.

Answer 2

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.