[proofplan] We first realize $A$ by a faithful finite-dimensional $*$-representation and then pass to an irreducible subrepresentation. Since the kernel of a nonzero $*$-representation is a closed two-sided ideal, simplicity forces this irreducible summand to remain faithful. Burnside's theorem then identifies the image of the irreducible representation with the full operator algebra on its [Hilbert space](/page/Hilbert%20Space). Choosing a basis of that finite-dimensional Hilbert space identifies the operator algebra with a full matrix algebra. [/proofplan] [step:Choose a faithful finite-dimensional representation and an irreducible summand] First note that $A$ is unital: this follows from the finite-dimensional $C^*$-algebra structure theorem [citetheorem:8588], which expresses $A$ as a finite direct sum of full matrix algebras and hence gives a multiplicative identity because $A\neq 0$. For any complex Hilbert space $E$, write $\mathcal L(E)$ for the algebra of bounded linear operators from $E$ to itself. The standard finite-dimensional representation theorem for unital finite-dimensional $C^*$-algebras gives a finite-dimensional complex Hilbert space $K$ and an injective unital $*$-homomorphism \begin{align*} \Pi:A\to \mathcal L(K). \end{align*} Because $A\neq 0$ and $\Pi$ is injective, $\Pi$ is not the zero representation, so $K\neq \{0\}$. Among all nonzero closed subspaces of $K$ invariant under $\Pi(a)$ for every $a\in A$, choose one of minimal dimension, and denote it by $H$. Since $K$ is finite-dimensional, such a subspace exists. For every $a\in A$, the operator $\Pi(a)$ maps $H$ into $H$, so define \begin{align*} \pi:A \to \mathcal L(H), \qquad a \mapsto \Pi(a)|_H. \end{align*} The map $\pi$ is a $*$-representation. Restriction preserves addition, scalar multiplication, and multiplication because $H$ is invariant under every $\Pi(a)$. To check the adjoint operation, fix $a\in A$. Since $a^*\in A$, invariance gives $\Pi(a^*)H\subset H$; because $\Pi$ is a $*$-homomorphism, $\Pi(a^*)=\Pi(a)^*$. Hence $\Pi(a)^*|_H$ is an operator on $H$, and the adjoint of $\Pi(a)|_H$ in $\mathcal L(H)$ is precisely $\Pi(a)^*|_H=\Pi(a^*)|_H$. Therefore $\pi(a)^*=\pi(a^*)$. The minimality of $H$ implies that $\pi$ is irreducible: if $V\subset H$ is a nonzero closed subspace invariant under every $\pi(a)$, then $V$ is a nonzero $\Pi(A)$-invariant subspace of $K$ contained in $H$, hence $V=H$. [guided] We want a representation whose image is as large as possible. Burnside's theorem applies to irreducible finite-dimensional operator algebras, so the first task is to pass from a faithful representation to an irreducible one. As above, $A$ is unital by the finite-dimensional $C^*$-algebra structure theorem [citetheorem:8588], and $\mathcal L(E)$ denotes the algebra of bounded linear operators from a complex Hilbert space $E$ to itself. The standard finite-dimensional representation theorem for unital finite-dimensional $C^*$-algebras gives a finite-dimensional complex Hilbert space $K$ and an injective unital $*$-homomorphism \begin{align*} \Pi:A\to \mathcal L(K). \end{align*} This is faithful because injectivity means that different elements of $A$ act as different operators on $K$. Since $A\neq 0$, the image $\Pi(A)$ is not the zero algebra, and hence $K\neq \{0\}$. Now consider the collection of all nonzero closed subspaces $E\subset K$ such that $\Pi(a)E\subset E$ for every $a\in A$. The space $K$ itself belongs to this collection, and because $K$ is finite-dimensional there is an element of this collection with minimal dimension. Fix such a subspace and call it $H$. For each $a\in A$, define \begin{align*} \pi:A&\to \mathcal L(H) \end{align*} \begin{align*} a&\mapsto \Pi(a)|_H. \end{align*} This map is well-defined because $H$ is invariant under every $\Pi(a)$. It is a $*$-representation. Linearity and multiplicativity follow from restricting the corresponding identities for $\Pi$ to $H$. The only point requiring care is the adjoint: fix $a\in A$. Since $A$ is closed under the involution, $a^*\in A$, and the defining invariance of $H$ gives $\Pi(a^*)H\subset H$. Because $\Pi$ is a $*$-homomorphism, $\Pi(a^*)=\Pi(a)^*$. Thus $\Pi(a)^*$ maps $H$ into $H$, so the Hilbert-space adjoint of $\Pi(a)|_H$ is $\Pi(a)^*|_H=\Pi(a^*)|_H$. Therefore $\pi(a)^*=\pi(a^*)$. This proves that restriction preserves the involution, so $\pi$ is a $*$-representation. Finally, $\pi$ is irreducible. Indeed, if $V\subset H$ is a nonzero closed subspace invariant under $\pi(a)$ for every $a\in A$, then $V$ is also invariant under $\Pi(a)$ for every $a\in A$. Since $V\subset H$ and $H$ was chosen with minimal positive dimension among invariant subspaces, we must have $V=H$. Thus $H$ has no proper nonzero invariant subspaces for the representation $\pi$. [/guided] [/step] [step:Use simplicity to prove the irreducible summand is faithful] The kernel \begin{align*} \ker\pi:=\{a\in A:\pi(a)=0\} \end{align*} is a closed two-sided ideal of $A$, because $\pi$ is a continuous algebra homomorphism. Since $A$ is simple and nonzero, its only closed two-sided ideals are $\{0\}$ and $A$. The representation $\pi$ is nonzero: since $H\neq\{0\}$ and $\Pi$ was chosen unital, $\pi(1_A)=\operatorname{id}_H\neq 0$, where $1_A$ denotes the multiplicative identity of $A$ and $\operatorname{id}_H$ denotes the identity operator on $H$. Hence $\ker\pi\neq A$. Therefore simplicity gives \begin{align*} \ker\pi=\{0\}. \end{align*} Thus $\pi$ is faithful. [/step] [step:Apply Burnside's theorem to identify the image with all operators on $H$] Let \begin{align*} B:=\pi(A)\subset \mathcal L(H). \end{align*} Then $B$ is a unital complex $*$-subalgebra of $\mathcal L(H)$. The space $H$ is finite-dimensional and nonzero, and $B$ acts irreducibly on $H$ because $\pi$ is irreducible. By Burnside's theorem for finite-dimensional irreducible complex operator algebras, an irreducible unital complex subalgebra of $\mathcal L(H)$ is all of $\mathcal L(H)$. Therefore \begin{align*} \pi(A)=\mathcal L(H). \end{align*} Since $\pi$ is injective and surjective onto $\mathcal L(H)$, it is a $*$-isomorphism \begin{align*} A\cong \mathcal L(H). \end{align*} [/step] [step:Choose a basis of $H$ and identify $\mathcal L(H)$ with a matrix algebra] Set \begin{align*} n:=\dim_{\mathbb C} H. \end{align*} Because $H\neq \{0\}$, we have $n\in\mathbb N$. Choose an ordered [orthonormal basis](/page/Orthonormal%20Basis) \begin{align*} \mathcal B=(e_1,\dots,e_n) \end{align*} of $H$. Define \begin{align*} \Theta:\mathcal L(H)\to M_n(\mathbb C), \qquad T\mapsto [T]_{\mathcal B}. \end{align*} where $[T]_{\mathcal B}$ denotes the matrix of the operator $T$ with respect to the basis $\mathcal B$. The map $\Theta$ is a unital algebra isomorphism, preserves adjoints because the basis is orthonormal, and preserves the $C^*$-norm because it is a $*$-isomorphism between finite-dimensional concrete $C^*$-algebras. Combining the $*$-isomorphism $\pi:A\to\mathcal L(H)$ with $\Theta$ gives a $*$-isomorphism \begin{align*} \Theta\circ \pi:A\to M_n(\mathbb C). \end{align*} Therefore $A\cong M_n(\mathbb C)$ as $C^*$-algebras. [/step]