[proofplan] We use the tensor-norm characterization of nuclearity: a $C^*$-algebra $A$ is nuclear precisely when, for every $C^*$-algebra $B$, the minimal and maximal $C^*$-norms on $A\odot B$ agree. For commutative algebras, the Gelfand representation reduces the problem to $C_0(X)$; the compact case follows from the representation of $C(X)\otimes B$ as continuous $B$-valued functions, and the locally compact case follows by unitization and passage to closed ideals. For finite-dimensional algebras, the structure theorem reduces the problem to finite direct sums of matrix algebras, and matrix algebras have a unique tensor norm because $M_n(\mathbb C)\odot B$ is canonically $M_n(B)$. [/proofplan] [step:Use tensor-norm uniqueness as the definition of nuclearity] Let $A$ and $B$ be $C^*$-algebras, and let $A\odot B$ denote their algebraic $*$-[tensor product](/page/Tensor%20Product). We write $\|\cdot\|_{\min}$ for the minimal $C^*$-tensor norm and $\|\cdot\|_{\max}$ for the maximal $C^*$-tensor norm on $A\odot B$. By the tensor-norm characterization of nuclearity, a $C^*$-algebra $A$ is nuclear iff for every $C^*$-algebra $B$ the canonical quotient map \begin{align*} A\otimes_{\max} B \to A\otimes_{\min} B \end{align*} is isometric, equivalently iff $A\odot B$ has only one $C^*$-cross norm extending the given norms on $A$ and $B$. We use this characterization throughout. [/step] [step:Prove that commutative $C^*$-algebras are nuclear] Let $A$ be a commutative $C^*$-algebra. By the commutative [Gelfand representation theorem](/theorems/2678) for $C^*$-algebras, there is a locally compact [Hausdorff space](/page/Hausdorff%20Space) $X$ and a $*$-isomorphism \begin{align*} A \cong C_0(X). \end{align*} Since nuclearity is preserved by $*$-isomorphism, it is enough to prove that $C_0(X)$ is nuclear. First suppose $X$ is compact. Let $B$ be any $C^*$-algebra, and let $z\in C(X)\odot B$. Under the standard identification \begin{align*} C(X)\odot B \subset C(X,B), \end{align*} the element $z$ is a [continuous function](/page/Continuous%20Function) $z:X\to B$ of the form \begin{align*} z(x)=\sum_{k=1}^{m} f_k(x)b_k \end{align*} for some $m\in\mathbb N$, functions $f_k\in C(X)$, and elements $b_k\in B$. Let $\gamma$ be any $C^*$-cross norm on $C(X)\odot B$, and let $D_\gamma$ be the $C^*$-completion. Let \begin{align*} \iota_X:C(X)\to D_\gamma \end{align*} and \begin{align*} \iota_B:B\to D_\gamma \end{align*} be the canonical $*$-homomorphisms determined by $f\mapsto f\otimes 1$ and $b\mapsto 1\otimes b$, using unitizations if $B$ is nonunital. Since $\gamma$ is a cross norm, these maps are isometric on the original factors, and their ranges commute. Choose a faithful representation \begin{align*} \pi:D_\gamma\to \mathcal{L}(H) \end{align*} on a complex [Hilbert space](/page/Hilbert%20Space) $H$. The representation $\pi\circ\iota_X$ of the unital commutative $C^*$-algebra $C(X)$ has a spectral measure $E$ on $X$. Because $\pi(\iota_B(B))$ commutes with $\pi(\iota_X(C(X)))$, it commutes with every spectral projection $E(S)$ for every Borel set $S\subset X$. Hence the spectral theorem gives a direct-integral decomposition of $H$ over $X$ in which $\pi(\iota_X(f))$ acts by multiplication by $f$ and each $\pi(\iota_B(b))$ acts fiberwise by a representation of $B$. Therefore $\pi(z)$ acts fiberwise as the operator \begin{align*} \sum_{k=1}^{m} f_k(x)\pi_x(b_k) \end{align*} for a measurable field of $*$-representations $\pi_x:B\to\mathcal{L}(H_x)$. Each $\pi_x$ is contractive by [citetheorem:8547], so for every $x\in X$, \begin{align*} \left\|\sum_{k=1}^{m} f_k(x)\pi_x(b_k)\right\|_{\mathcal{L}(H_x)}\le \left\|\sum_{k=1}^{m} f_k(x)b_k\right\|_B=\|z(x)\|_B. \end{align*} Taking the essential supremum of the fiber norms and using faithfulness of $\pi$ gives \begin{align*} \gamma(z)=\|\pi(z)\|_{\mathcal{L}(H)}\le \sup_{x\in X}\|z(x)\|_B. \end{align*} The right-hand side is exactly the spatial norm of $z$ under the standard identification $C(X)\otimes_{\min}B\cong C(X,B)$. Since the minimal norm is bounded above by every $C^*$-cross norm on the algebraic tensor product, we also have \begin{align*} \|z\|_{\min}\le \gamma(z). \end{align*} Thus $\gamma(z)=\|z\|_{\min}$ for every $z\in C(X)\odot B$. Since $B$ was arbitrary, $C(X)$ is nuclear. Now let $X$ be locally compact. Let $X^+$ denote the one-point compactification of $X$. The unitization of $C_0(X)$ is canonically $C(X^+)$, and the inclusion \begin{align*} C_0(X)\trianglelefteq C(X^+) \end{align*} identifies $C_0(X)$ with a closed two-sided ideal. The compact case gives that $C(X^+)$ is nuclear. By the permanence theorem that closed two-sided ideals of nuclear $C^*$-algebras are nuclear, $C_0(X)$ is nuclear. Hence $A$ is nuclear. [/step] [step:Prove that full matrix algebras are nuclear] Let $n\in\mathbb N$. We prove that $M_n(\mathbb C)$ is nuclear. Let $B$ be a $C^*$-algebra. The algebraic map \begin{align*} \Theta:M_n(\mathbb C)\odot B\to M_n(B) \end{align*} defined on elementary tensors by \begin{align*} \Theta(e_{ij}\otimes b)=(b)_{ij} \end{align*} is a $*$-isomorphism, where $e_{ij}\in M_n(\mathbb C)$ is the standard matrix unit and $(b)_{ij}\in M_n(B)$ denotes the matrix with $b$ in the $(i,j)$ entry and $0$ elsewhere. The $C^*$-norm on the algebraic $*$-algebra $M_n(B)$ is unique. Indeed, let $\gamma$ be any $C^*$-norm on $M_n(B)$, and let $D_\gamma$ be its completion. The images of the constant matrices $e_{ij}$ form a system of matrix units in $D_\gamma$. Let \begin{align*} C_\gamma:=e_{11}D_\gamma e_{11} \end{align*} be the first corner. The map $b\mapsto e_{11}be_{11}$ identifies $B$ algebraically with a dense $*$-subalgebra of $C_\gamma$, and the restricted norm is a $C^*$-norm on the original $C^*$-algebra $B$. By the uniqueness of the $C^*$-norm [citetheorem:8546], this restricted norm is exactly $\|\cdot\|_B$, so $C_\gamma$ is the usual completion of $B$. The matrix units give a $*$-isomorphism \begin{align*} D_\gamma\to M_n(C_\gamma) \end{align*} by sending $x\in D_\gamma$ to the matrix whose $(i,j)$ entry is $e_{1i}xe_{j1}\in C_\gamma$. Its inverse sends $(c_{ij})$ to $\sum_{i,j=1}^{n}e_{i1}c_{ij}e_{1j}$. These formulas are inverse $*$-homomorphisms because $e_{ij}e_{kl}=\delta_{jk}e_{il}$. Thus $D_\gamma$ is $*$-isomorphic to the standard $C^*$-algebra $M_n(B)$. Applying [citetheorem:8546] to the identity algebra $M_n(B)$ with the usual norm and the norm induced by $\gamma$ shows that $\gamma$ is the usual $C^*$-norm on $M_n(B)$. Consequently every $C^*$-norm on $M_n(\mathbb C)\odot B$ agrees with the norm transported from $M_n(B)$ through $\Theta$. Since $B$ was arbitrary, $M_n(\mathbb C)$ is nuclear. [guided] We fix $n\in\mathbb N$ and prove that $M_n(\mathbb C)$ has a unique tensor norm against every $C^*$-algebra $B$. The relevant algebraic identification is the map \begin{align*} \Theta:M_n(\mathbb C)\odot B\to M_n(B) \end{align*} defined by \begin{align*} \Theta(e_{ij}\otimes b)=(b)_{ij}. \end{align*} Here $e_{ij}$ is the standard matrix unit in $M_n(\mathbb C)$, and $(b)_{ij}$ is the $n\times n$ matrix over $B$ with entry $b$ in position $(i,j)$ and zero in all other positions. This map preserves multiplication because matrix units satisfy \begin{align*} e_{ij}e_{kl}=\delta_{jk}e_{il}, \end{align*} and it preserves the involution because \begin{align*} (e_{ij}\otimes b)^*=e_{ji}\otimes b^*. \end{align*} Thus $\Theta$ is a $*$-isomorphism of algebraic $*$-algebras. It remains to explain why $M_n(B)$ has only one $C^*$-norm. Let $\gamma$ be a $C^*$-norm on the algebraic $*$-algebra $M_n(B)$, and let $D_\gamma$ be its $C^*$-completion. The diagonal matrix units define projections in $D_\gamma$, and the full family $(e_{ij})_{1\le i,j\le n}$ remains a system of matrix units because the multiplication and involution are the same algebraic operations. Consider the first corner \begin{align*} C_\gamma:=e_{11}D_\gamma e_{11}. \end{align*} The algebraic copy of $B$ in the $(1,1)$ entry is dense in this corner. The norm induced by $\gamma$ on that copy is a $C^*$-norm on the same involutive algebra $B$. Since $B$ already has its given $C^*$-norm, the uniqueness of the $C^*$-norm [citetheorem:8546] implies that the induced norm on the corner copy is exactly $\|\cdot\|_B$. Thus the corner completion $C_\gamma$ is the original $C^*$-algebra $B$. Now the matrix units reconstruct the whole completion from this first corner. Define \begin{align*} \Psi:D_\gamma\to M_n(C_\gamma) \end{align*} by declaring that the $(i,j)$ entry of $\Psi(x)$ is $e_{1i}xe_{j1}\in C_\gamma$. Define \begin{align*} \Omega:M_n(C_\gamma)\to D_\gamma \end{align*} by \begin{align*} \Omega((c_{ij}))=\sum_{i=1}^{n}\sum_{j=1}^{n}e_{i1}c_{ij}e_{1j}. \end{align*} The relations $e_{ij}e_{kl}=\delta_{jk}e_{il}$ show that $\Psi$ and $\Omega$ preserve multiplication and involution and are inverse to each other. Hence $D_\gamma$ is $*$-isomorphic to $M_n(C_\gamma)=M_n(B)$ with its standard $C^*$-norm. Applying [citetheorem:8546] to the algebraic $*$-algebra $M_n(B)$ with the usual norm and the norm induced by $\gamma$ gives equality of the two norms. Transporting this unique norm back along $\Theta$ shows that every $C^*$-cross norm on $M_n(\mathbb C)\odot B$ is the same. Since this holds for every $C^*$-algebra $B$, $M_n(\mathbb C)$ is nuclear. [/guided] [/step] [step:Pass from matrix blocks to finite-dimensional $C^*$-algebras] Let $A$ be a finite-dimensional $C^*$-algebra. By [citetheorem:8588], there exist integers $r\ge 0$ and $n_1,\dots,n_r\in\mathbb N$ such that \begin{align*} A\cong \bigoplus_{j=1}^{r} M_{n_j}(\mathbb C) \end{align*} as $C^*$-algebras, with the convention that $r=0$ means $A=0$. For each $j\in\{1,\dots,r\}$, the preceding step proves that $M_{n_j}(\mathbb C)$ is nuclear. We now use that finite direct sums preserve nuclearity. To see this in the tensor-norm formulation, let $B$ be a $C^*$-algebra. Algebraically, \begin{align*} \left(\bigoplus_{j=1}^{r}M_{n_j}(\mathbb C)\right)\odot B \cong \bigoplus_{j=1}^{r}\left(M_{n_j}(\mathbb C)\odot B\right). \end{align*} The central projections of the finite direct sum split every $C^*$-completion into the corresponding closed summands, and the norm of an element is the maximum of the norms of its summand components. Since each summand has a unique $C^*$-tensor norm, the whole finite direct sum has a unique $C^*$-tensor norm. Therefore $\bigoplus_{j=1}^{r}M_{n_j}(\mathbb C)$ is nuclear. Nuclearity is preserved under $*$-isomorphism, so $A$ is nuclear. This proves the finite-dimensional assertion, and together with the commutative case completes the proof. [/step]