[proofplan] We first separate the zero algebra, since the remaining argument uses the unit of $A$. For nonzero finite-dimensional $A$, the centre is a finite-dimensional commutative $C^*$-algebra, so its minimal central projections split $A$ into finitely many central ideals. Each central summand has centre equal to the scalar multiples of its unit, which forces it to be simple. The finite-dimensional simple classification then identifies the summands with full matrix algebras, and uniqueness follows by comparing minimal central projections and matrix dimensions. [/proofplan] [step:Handle the zero algebra before using a unit] If $A=0$, take $r=0$. The empty direct sum is the zero $C^*$-algebra, so the asserted decomposition holds. The uniqueness assertion is only stated for $A\neq 0$, so there is nothing further to prove in this case. For the rest of the proof, assume $A\neq 0$. [/step] [step:Use the centre to obtain minimal central projections] We use the standard finite-dimensional fact that every nonzero finite-dimensional $C^*$-algebra is unital (citing a result not yet in the wiki: finite-dimensional $C^*$-algebras are unital when nonzero). Let $1_A$ denote its unit. Define the centre of $A$ by \begin{align*} Z(A):=\{z\in A: za=az \text{ for every } a\in A\}. \end{align*} Then $Z(A)$ is a finite-dimensional commutative unital $C^*$-subalgebra of $A$. By the finite-dimensional commutative $C^*$-algebra classification (citing a result not yet in the wiki: finite-dimensional commutative $C^*$-algebras are isomorphic to $\mathbb C^r$), there are an integer $r\in\mathbb N$ and a unital $*$-isomorphism \begin{align*} \theta:Z(A)\to \mathbb C^r. \end{align*} For each $j\in\{1,\dots,r\}$, let $e_j\in \mathbb C^r$ be the coordinate unit with $1$ in the $j$th coordinate and $0$ elsewhere, and define \begin{align*} p_j:=\theta^{-1}(e_j)\in Z(A). \end{align*} Since each $e_j$ is a projection in $\mathbb C^r$ and $\theta$ is a $*$-isomorphism, each $p_j$ is a projection in $Z(A)$. Since $e_i e_j=0$ for $i\neq j$ and $\sum_{j=1}^r e_j=(1,\dots,1)$, we have \begin{align*} p_i p_j=0 \quad \text{for } i\neq j \end{align*} and \begin{align*} 1_A=\sum_{j=1}^r p_j. \end{align*} Moreover, the projections $p_1,\dots,p_r$ are precisely the minimal nonzero central projections: if $q\in Z(A)$ is a nonzero projection with $q\le p_j$, then $\theta(q)$ is a nonzero projection in the one-dimensional algebra $\mathbb C e_j$, hence $\theta(q)=e_j$, so $q=p_j$. [/step] [step:Split $A$ as the direct sum of the central ideals $p_jA$] For each $j\in\{1,\dots,r\}$, define \begin{align*} A_j:=p_jA=\{p_j a:a\in A\}. \end{align*} Because $p_j$ is central, $A_j$ is a two-sided $*$-ideal of $A$, and $p_j$ is the unit of $A_j$. Define the map \begin{align*} \Phi:A\to \bigoplus_{j=1}^r A_j \end{align*} by \begin{align*} \Phi(a):=(p_1a,\dots,p_ra). \end{align*} Define also \begin{align*} \Psi:\bigoplus_{j=1}^r A_j\to A \end{align*} by \begin{align*} \Psi(a_1,\dots,a_r):=\sum_{j=1}^r a_j. \end{align*} The orthogonality relations $p_i p_j=0$ for $i\neq j$ show that $\Phi$ and $\Psi$ are inverse algebra homomorphisms. Since each $p_j$ is self-adjoint and central, both maps preserve the adjoint operation. Thus $\Phi$ is a bijective $*$-homomorphism. By [citetheorem:8547], applied to $\Phi$ and to its inverse $\Psi$, the map $\Phi$ is isometric. Hence \begin{align*} A\cong \bigoplus_{j=1}^r A_j \end{align*} as $C^*$-algebras. [/step] [step:Show each central summand is simple] Fix $j\in\{1,\dots,r\}$, and write $B:=A_j=p_jA$. We first identify the centre of $B$. Let \begin{align*} Z(B):=\{b\in B: bx=xb \text{ for every } x\in B\}. \end{align*} If $z\in Z(B)$ and $a\in A$, then \begin{align*} a=\sum_{k=1}^r p_k a. \end{align*} Since $z\in p_jA$, we have $zp_k=0=p_kz$ for $k\neq j$, and since $z$ is central in $B$, it commutes with $p_j a\in B$. Therefore $za=az$ for every $a\in A$, so $z\in Z(A)$. Also $p_jz=z$, so $z$ lies in the one-dimensional central summand $\mathbb C p_j$. Hence \begin{align*} Z(B)=\mathbb C p_j. \end{align*} Now let $I\trianglelefteq B$ be a nonzero closed two-sided ideal. Since $I$ is a nonzero finite-dimensional $C^*$-algebra, it has a unit $q\in I$ (again using the standard finite-dimensional unitality fact for nonzero $C^*$-algebras). The element $q$ is a projection. For every $b\in B$, both $qb$ and $bq$ belong to $I$, and the identity property of $q$ on $I$ gives \begin{align*} (qb)q=qb \end{align*} and \begin{align*} q(bq)=bq. \end{align*} Thus $qb=q b q=bq$, so $q\in Z(B)$. Since $Z(B)=\mathbb C p_j$ and $q$ is a nonzero projection, we must have $q=p_j$. Therefore $p_j\in I$, and since $p_j$ is the unit of $B$, it follows that $I=B$. Thus $B=A_j$ is simple. [guided] Fix $j\in\{1,\dots,r\}$ and set $B:=p_jA$. The point is to prove that the central projection $p_j$ cuts out an indecomposable summand. We do this in two stages: first compute the centre of $B$, then use the identity of any nonzero ideal to manufacture a central projection. Define \begin{align*} Z(B):=\{b\in B: bx=xb \text{ for every } x\in B\}. \end{align*} We claim that $Z(B)=\mathbb C p_j$. Let $z\in Z(B)$. Since $1_A=\sum_{k=1}^r p_k$, every $a\in A$ decomposes as \begin{align*} a=\sum_{k=1}^r p_k a. \end{align*} Because $z\in B=p_jA$, we have $zp_k=0=p_kz$ whenever $k\neq j$. The only remaining component is $p_j a$, which belongs to $B$, and $z$ commutes with every element of $B$ by the definition of $Z(B)$. Therefore $z$ commutes with each summand $p_k a$, and hence $za=az$ for every $a\in A$. This proves $z\in Z(A)$. We also have $p_jz=z$, so $z$ is supported on the central projection $p_j$. Under the isomorphism $\theta:Z(A)\to\mathbb C^r$, this says that $\theta(z)$ is supported only in the $j$th coordinate. The elements of $\mathbb C^r$ supported only in the $j$th coordinate are exactly the scalar multiples of $e_j$. Pulling back by $\theta$, we obtain $z\in\mathbb C p_j$. Since every scalar multiple of $p_j$ plainly commutes with $B$, this proves \begin{align*} Z(B)=\mathbb C p_j. \end{align*} Now let $I\trianglelefteq B$ be a nonzero closed two-sided ideal. Because $I$ is finite-dimensional and is itself a nonzero $C^*$-algebra, the standard finite-dimensional unitality fact gives a unit $q\in I$. This unit is a projection. We show that it is central in $B$. For any $b\in B$, the two-sided ideal property gives $qb\in I$ and $bq\in I$. Since $q$ is the identity element of $I$, multiplication by $q$ fixes these two elements: \begin{align*} (qb)q=qb \end{align*} and \begin{align*} q(bq)=bq. \end{align*} Both expressions have the same middle product $q b q$, so $qb=bq$. Since this holds for every $b\in B$, we have $q\in Z(B)$. The equality $Z(B)=\mathbb C p_j$ now forces $q=\lambda p_j$ for some $\lambda\in\mathbb C$. Because $q$ is a nonzero projection, the scalar $\lambda$ satisfies $\lambda^2=\lambda$ and $\lambda\neq 0$, hence $\lambda=1$. Therefore $q=p_j$. Since $p_j$ is the unit of $B$ and $p_j\in I$, every $b\in B$ satisfies \begin{align*} b=p_jb\in I. \end{align*} Thus $I=B$. We have proved that $B=p_jA$ has no nonzero proper closed two-sided ideals, so $B$ is simple. [/guided] [/step] [step:Identify the simple summands with full matrix algebras] For each $j\in\{1,\dots,r\}$, the algebra $A_j=p_jA$ is a finite-dimensional simple $C^*$-algebra. By [citetheorem:8587], there exists an integer $n_j\in\mathbb N$ and a $C^*$-algebra isomorphism \begin{align*} A_j\cong M_{n_j}(\mathbb C). \end{align*} Combining these isomorphisms with the direct-sum decomposition of $A$ gives \begin{align*} A\cong \bigoplus_{j=1}^r M_{n_j}(\mathbb C) \end{align*} as $C^*$-algebras. [/step] [step:Recover the multiset of matrix sizes from the algebra] Assume $A\neq 0$ and suppose also that \begin{align*} A\cong \bigoplus_{k=1}^s M_{m_k}(\mathbb C) \end{align*} as $C^*$-algebras, with $m_1,\dots,m_s\in\mathbb N$. The centre of $\bigoplus_{j=1}^r M_{n_j}(\mathbb C)$ is \begin{align*} \bigoplus_{j=1}^r \mathbb C 1_{M_{n_j}(\mathbb C)}, \end{align*} and its minimal nonzero central projections are exactly the coordinate units. The same description holds for the second decomposition with $s$ summands. A $*$-isomorphism maps centres to centres and maps minimal nonzero central projections to minimal nonzero central projections. Hence it gives a bijection between the $r$ coordinate central projections in the first decomposition and the $s$ coordinate central projections in the second decomposition. Therefore $r=s$, and after reindexing there are $*$-isomorphisms \begin{align*} M_{n_j}(\mathbb C)\cong M_{m_j}(\mathbb C) \end{align*} for each $j\in\{1,\dots,r\}$. A $*$-isomorphism is in particular a complex-linear vector-space isomorphism. Since \begin{align*} \dim_{\mathbb C} M_n(\mathbb C)=n^2 \end{align*} for every $n\in\mathbb N$, the equality of dimensions gives $n_j^2=m_j^2$. Because $n_j$ and $m_j$ are positive integers, $n_j=m_j$. Thus the multiset $\{n_1,\dots,n_r\}$ is uniquely determined by $A$. [/step]