Let $A$ be a $C^*$-algebra and $S\subset A$ be a nonempty set. From the definition, the $C^*$-algebra generated by $S$ is the smallest $C^*$-algebra in $A$ containing $S$. My question: is there a concrete form of the $C^*$-algebra generated by $S$?
For a group $G$, the subgroup generated by finite set $\{a_1,a_2,...,a_n\}$ will consists of all finite products $a_1^{n_1}a_2^{n_2},...,a_t^{n_t}$. How about the category of $C^*$-algebras?
Let's think about it for a moment. A $C^*$-subalgebra has to be closed under the algebra operations (sum, product and scalar multiplication), under conjugation, and since it has to be a Banach space also closed in norm.
So first we need to take the conjugates of elements of $S$, let's call this new set $S\cup S^*$. Then we need to take all linear combinations of finite products of elements of $S\cup S^*$. That will be a $*$-subalgebra, let's call it $B$. Finally, we take the closure of $B$, and that will give us the $C^*$-subalgebra generated by $S$. (of course you need to check that the closure is still a subalgebra and closed under conjugation, but that's not difficult to see, as all operations are continuous) So a general element is a limit of a sequence of elements of the form:
$\sum_{i=1}^n \lambda_i s_{i,1}s_{i,2}...s_{i, k_i}$
Where $n\geq 0$, $s_{i,j}$ are elements in $S\cup S^*$ and $\lambda_i\in\mathbb{C}$. Not a very nice description, but it is the best we can get. A $C^*$-algebra is a much more complicated structure than a group or vector space.