# Irreducible *-representations of $M_n(\mathbb{C})$

This might be a very stupid question but, is the defining representation the only irreducible *-representation of $$M_n(\mathbb{C})$$? This seems to me that this should be the case but I would have no idea how to go around and prove it.

I just realized this may be a simple consequence of the fact that a non-degenerate representation of a $$C^*$$-algebra is irreducible if and only if the associated vector states induced are pure. Since the pure states are the vector states in the defining representation and the GNS representation obtained from them is once again the defining representation, we get that the only irreducible representation is the defining one. I would however prefer a simple proof which required the least amount of prerequisites. Thanks!

This answer is probably inaccurate since I don't know that "defining representation" is. If you mean the identity representation, then the answer is no: any representation of the form $$A\longmapsto UAU^*$$, for a unitary $$U$$, is irreducible and of course it changes matrices (it corresponds to choosing different orthonormal bases in $$\mathbb C^n$$).
But the above is all there is. Since $$M_n(\mathbb C)$$ is simple, any representation is faithful. One can then deduce that any representation $$\pi$$ is of the form $$A\longmapsto W(A\otimes I)W^*\subset \mathbb C^n\otimes H$$ for some unitary $$W$$. It follows that $$\pi$$ is irreducible precisely when $$\dim H=1$$.
The proof when we assume that $$\pi$$ is irreducible is fairly simple. Consider the canonical matrix units $$\{E_{kj}:\ k,j=1,\ldots,n\}$$ and $$\pi:M_n(\mathbb C)\to B(H_\pi)$$ irreducible. The set $$\mathcal A=\operatorname{span}\{\pi(E_{kj}):\ k,j=1,\ldots,n\}\subset B(H_\pi)$$ is a $$*$$-subalgebra of $$B(H_\pi)$$. Since $$\pi$$ is irreducible, it is dense; being finite-dimensional, it is closed. So $$B(H_\pi)=\pi(M_n(\mathbb C))$$. Now you can use the matrix units $$\pi(E_{kj})$$ to construct an orthonormal basis $$\{f_k\}$$ of $$H_\pi$$, with $$\pi(E_{kj})f_s=\delta_{s,j}\,f_k$$. Then the map $$W:e_j\longmapsto f_j$$ (where $$\{e_j\}$$ is the canonical basis of $$\mathbb C^n$$) is a unitary with $$WE_{kj}W^*=\pi(E_{kj})$$. It follows that $$\pi=W\cdot W^*$$.