Understand the notation of a unitization $C^\ast$-algebra

Question

I have the following $C^\ast$-algebra $\mathcal{A}$ with norm $\|\cdot\|$. Let $\tilde{\mathcal{A}}=\mathcal{A}\oplus \mathbb{C}$ as a vector space. We endow it with multiplication and involution,

$$(a,\lambda)\cdot (b,\mu):=(ab+\lambda b+\mu a,\lambda \mu)$$ and $$(a,\lambda)^\ast:=(a^\ast,\bar{\lambda})$$

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OK I want to show that this does make a $\ast$-algebra. I know the properties to verify this, i.e.

1. $(a^\ast)^\ast=a$,

2. $(\lambda a+b)^\ast = \overline{\lambda}a^\ast+b^\ast$,

3. $(ab)^\ast=b^\ast a^\ast$.

However I am confused to show it because I don't understand the notation. If I want to show the first, i.e. $(a^\ast)^\ast$ then I should let $(ab+\lambda b+\mu a,\lambda\mu)^\ast$ then compute $((ab+\lambda b+\mu a,\lambda\mu)^\ast)^\ast$? For me it seems wrong..

I don't get the notation so if anyone can help me I would appreciate it. If one would show i.e. the first, then I will try the second and third property.

Thanks in advance.

Answer 1

In $$(a,\lambda)^\ast:=(a^\ast,\bar{\lambda})$$ if you repalce $a$ by $a^{*}$ and $\lambda$ by $\overline\lambda$ you get $$(a^\ast,\bar{\lambda})^\ast=(a, \lambda)$$ by properteis of ${*}$ in $\mathcal A$. Combine these two equations to get $$((a,\lambda)^\ast)^{\ast}=(a^\ast,\bar{\lambda})^\ast=(a,\lambda).$$