universal $C^*$-algebra isomorphic to $K(l^2(I))$ Let $I\neq \emptyset$ be a set, $X=\{e_{ij}:i,j\in I\}$ with the relations $$R=\{e_{ij}e_{kl}=\begin{cases}e_{il},  & \text{if }j=k,\\ 0, & \text{ else} \end{cases},\;\; e_{ij}^*=e_{ji}:\; i,j,k,l\in I\}.$$
Prove that $C^*(X,R)\cong K(l^2(I))$. Here is $C^*(X,R)$ the universal $C^*$-algebra for $(X,R)$ and $K(l^2(I))$ are the linear, bounded operators $l^2(I)\to l^2(I)$ which are compact.
The strategy is to prove first that $C^*(X,R)$ exists. I've already done that. Then I've constructed a representation $\varphi:X\to K(l^2(I)), e_{ij}\mapsto E_{ij}$ (i.e. $\varphi$ preserves the relations) where $\{E_{ij} \}_{i,j\in I}$ are the rank-1 projections in $K(l^2(I))$.
Then from lecture we know that $\varphi$ induces a $*$-homomorphism $\hat{\varphi}:C^*(X,R)\to K(l^2(I))$. 
Now my questions: There is a hint on the worksheet that we have to show that every irreducible $\ast$-representation $\tau:C^*(X,R)\to L(H)$ is equivalent to $\hat{\varphi}$, i.e. there is an unitary operator $V:H\to l^2(I)$ such that $V\tau(a)=\hat{\varphi}(a) V$ for all $a\in A$. I know how to prove it, but why do I have to prove it? 
And why is $\hat{\varphi}$ bijective?
 A: You should prove that the $E_{ij}$ within $K(\ell^2(I))$ satisfy the universal property. This proves existence and uniqueness of $C^\ast(X;R)$ in one swoop.
Obviously the $E_{ij}$ are linearly independent, so whenever you have elements $x_{ij} \in A$ satisfying the relations, you can define a linear map by sending $E_{ij} \mapsto x_{ij}$. It is easy to see that this map will be compatible with multiplication and $\ast$ on the $\ast$-subalgebra $\mathcal{E}\subseteq L(\ell^2(I))$ generated by $\{E_{ij}\}$ (= the set of $I\times I$-matrices with only finitely many nonzero entries). What is left to show is the continuity of the map, so that this density can be used to uniquely extend it to the closure of $\mathcal{E}$ (which is $K(\ell^2(I))$ ). The extension will also be a $\ast$-algebra homomorphism.
EDIT: Rather than showing continuity directly, first note that continuity is automatic for finite $I$ because then $\mathcal{E} = K(\ell^2(I)) = L(\ell^2(I)) = \mathbb{C}^{I\times I}$ is finite-dimensional.
For the general case use that every element of $\mathcal{E}$ lies in the image of a suitable embedding $\mathbb{C}^{I_0\times I_0} \hookrightarrow L(\ell^2(I))$ for some finite $I_0\subseteq I$. This embedding already is a $C^\ast$-algebrahomomorphism so you know something about the norms. Put it all together and you get an upper bound for the norm of the linear map $\mathcal{E}\to A$.
