[proofplan] We first realize both matrix algebras on their standard Hilbert spaces and identify the Hilbert-space [tensor product](/page/Tensor%20Product) $\mathbb C^m\otimes \mathbb C^n$ with $\mathbb C^{mn}$ using the ordered basis indexed by pairs $(i,k)$. Under this identification, the elementary tensor $e_{ij}\otimes f_{kl}$ acts as the corresponding matrix unit $E_{(i,k),(j,l)}$. We then check multiplication and adjoints on matrix units, which proves that the resulting [linear map](/page/Linear%20Map) is a unital $*$-homomorphism. Finally, because the images are all matrix units of $M_{mn}(\mathbb C)$ and the minimal norm is the operator norm coming from this faithful spatial representation, the algebraic $*$-isomorphism extends to the claimed $C^*$-isomorphism. [/proofplan]

[step:Represent the matrix algebras on the standard tensor-product Hilbert space] Let $H_m:=\mathbb C^m$ with standard ordered basis $\{u_i:1\le i\le m\}$, and let $H_n:=\mathbb C^n$ with standard ordered basis $\{v_k:1\le k\le n\}$. The standard representations are the unital faithful $*$-homomorphisms \begin{align*} \pi_m:M_m(\mathbb C)\to \mathcal L(H_m) \end{align*} and \begin{align*} \pi_n:M_n(\mathbb C)\to \mathcal L(H_n) \end{align*} given by ordinary matrix multiplication. Define the ordered basis $\mathcal B$ of $H_m\otimes H_n$ by \begin{align*} \mathcal B:=\{u_i\otimes v_k:1\le i\le m,\ 1\le k\le n\} \end{align*} with lexicographic ordering on the pairs $(i,k)$. Let \begin{align*} U:H_m\otimes H_n\to \mathbb C^{mn} \end{align*} be the unitary map determined by sending $u_i\otimes v_k$ to the standard basis vector indexed by $(i,k)$. For $1\le i,j\le m$ and $1\le k,l\le n$, the spatial operator $(\pi_m(e_{ij})\otimes \pi_n(f_{kl}))$ satisfies \begin{align*} (\pi_m(e_{ij})\otimes \pi_n(f_{kl}))(u_r\otimes v_s)=\delta_{jr}\delta_{ls}\,u_i\otimes v_k \end{align*} for all $1\le r\le m$ and $1\le s\le n$. Therefore conjugation by $U$ sends this operator to the matrix unit $E_{(i,k),(j,l)}$ in $M_{mn}(\mathbb C)$. [/step]

[step:Define the algebraic map and verify multiplication on matrix units]Define the linear map \begin{align*} \Phi_0:M_m(\mathbb C)\odot M_n(\mathbb C)\to M_{mn}(\mathbb C) \end{align*} on the algebraic tensor product by \begin{align*} \Phi_0(e_{ij}\otimes f_{kl})=E_{(i,k),(j,l)}. \end{align*} The elementary tensors $e_{ij}\otimes f_{kl}$ form a basis of $M_m(\mathbb C)\odot M_n(\mathbb C)$, so this prescription defines a unique linear map. Let $1\le i,j,p,q\le m$ and $1\le k,l,r,s\le n$. Using the multiplication rule for matrix units, \begin{align*} (e_{ij}\otimes f_{kl})(e_{pq}\otimes f_{rs})=\delta_{jp}\delta_{lr}\,e_{iq}\otimes f_{ks}. \end{align*} Applying $\Phi_0$ gives \begin{align*} \Phi_0((e_{ij}\otimes f_{kl})(e_{pq}\otimes f_{rs}))=\delta_{jp}\delta_{lr}\,E_{(i,k),(q,s)}. \end{align*} On the other hand, \begin{align*} \Phi_0(e_{ij}\otimes f_{kl})\Phi_0(e_{pq}\otimes f_{rs})=E_{(i,k),(j,l)}E_{(p,r),(q,s)}. \end{align*} The multiplication rule in $M_{mn}(\mathbb C)$ gives \begin{align*} E_{(i,k),(j,l)}E_{(p,r),(q,s)}=\delta_{(j,l),(p,r)}E_{(i,k),(q,s)}. \end{align*} Since $\delta_{(j,l),(p,r)}=\delta_{jp}\delta_{lr}$, the two expressions agree. By bilinearity, $\Phi_0$ is multiplicative on all of $M_m(\mathbb C)\odot M_n(\mathbb C)$.[/step]

[guided]The point of checking multiplication only on matrix-unit tensors is that these tensors form a basis of the algebraic tensor product. Define \begin{align*} \Phi_0:M_m(\mathbb C)\odot M_n(\mathbb C)\to M_{mn}(\mathbb C) \end{align*} by requiring \begin{align*} \Phi_0(e_{ij}\otimes f_{kl})=E_{(i,k),(j,l)}. \end{align*} Because every element of $M_m(\mathbb C)\odot M_n(\mathbb C)$ is a unique finite linear combination of the tensors $e_{ij}\otimes f_{kl}$, this defines one and only one linear map. Now take two basis tensors, say $e_{ij}\otimes f_{kl}$ and $e_{pq}\otimes f_{rs}$. The multiplication in the algebraic tensor product is defined componentwise, so \begin{align*} (e_{ij}\otimes f_{kl})(e_{pq}\otimes f_{rs})=(e_{ij}e_{pq})\otimes(f_{kl}f_{rs}). \end{align*} The matrix-unit identities in the two factors are \begin{align*} e_{ij}e_{pq}=\delta_{jp}e_{iq} \end{align*} and \begin{align*} f_{kl}f_{rs}=\delta_{lr}f_{ks}. \end{align*} Therefore \begin{align*} (e_{ij}\otimes f_{kl})(e_{pq}\otimes f_{rs})=\delta_{jp}\delta_{lr}\,e_{iq}\otimes f_{ks}. \end{align*} Applying $\Phi_0$ gives \begin{align*} \Phi_0((e_{ij}\otimes f_{kl})(e_{pq}\otimes f_{rs}))=\delta_{jp}\delta_{lr}\,E_{(i,k),(q,s)}. \end{align*} We compare this with multiplying after applying $\Phi_0$. The images are matrix units in $M_{mn}(\mathbb C)$ indexed by ordered pairs, so \begin{align*} \Phi_0(e_{ij}\otimes f_{kl})\Phi_0(e_{pq}\otimes f_{rs})=E_{(i,k),(j,l)}E_{(p,r),(q,s)}. \end{align*} The product of these matrix units is nonzero exactly when the column index $(j,l)$ equals the row index $(p,r)$. Thus \begin{align*} E_{(i,k),(j,l)}E_{(p,r),(q,s)}=\delta_{(j,l),(p,r)}E_{(i,k),(q,s)}. \end{align*} Since equality of ordered pairs means both coordinates are equal, $\delta_{(j,l),(p,r)}=\delta_{jp}\delta_{lr}$. Hence multiplication before applying $\Phi_0$ gives the same result as multiplication after applying $\Phi_0$ on all basis tensors. Extending by bilinearity proves that $\Phi_0$ is multiplicative on the whole algebraic tensor product.[/guided]

[step:Verify that the algebraic map preserves adjoints and the unit] For matrix units, the involution rules are $e_{ij}^*=e_{ji}$, $f_{kl}^*=f_{lk}$, and $E_{(i,k),(j,l)}^*=E_{(j,l),(i,k)}$. Hence \begin{align*} \Phi_0((e_{ij}\otimes f_{kl})^*)=\Phi_0(e_{ji}\otimes f_{lk})=E_{(j,l),(i,k)}. \end{align*} Also, \begin{align*} \Phi_0(e_{ij}\otimes f_{kl})^*=E_{(i,k),(j,l)}^*=E_{(j,l),(i,k)}. \end{align*} Thus $\Phi_0(x^*)=\Phi_0(x)^*$ for every $x\in M_m(\mathbb C)\odot M_n(\mathbb C)$ by conjugate-linearity of the involution and linearity of $\Phi_0$. The unit of $M_m(\mathbb C)\odot M_n(\mathbb C)$ is \begin{align*} 1_{M_m(\mathbb C)}\otimes 1_{M_n(\mathbb C)}=\sum_{i=1}^m\sum_{k=1}^n e_{ii}\otimes f_{kk}. \end{align*} Therefore \begin{align*} \Phi_0(1_{M_m(\mathbb C)}\otimes 1_{M_n(\mathbb C)})=\sum_{i=1}^m\sum_{k=1}^n E_{(i,k),(i,k)}=1_{M_{mn}(\mathbb C)}. \end{align*} So $\Phi_0$ is a unital $*$-homomorphism. [/step]

[step:Show that the algebraic map is bijective] The set \begin{align*} \{E_{(i,k),(j,l)}:1\le i,j\le m,\ 1\le k,l\le n\} \end{align*} is the full system of matrix units of $M_{mn}(\mathbb C)$ relative to the ordered basis $\mathcal B$. Each of these matrix units is in the range of $\Phi_0$ because \begin{align*} E_{(i,k),(j,l)}=\Phi_0(e_{ij}\otimes f_{kl}). \end{align*} Since the matrix units form a basis of $M_{mn}(\mathbb C)$, $\Phi_0$ is surjective. The domain and codomain have the same complex dimension: \begin{align*} \dim_{\mathbb C}(M_m(\mathbb C)\odot M_n(\mathbb C))=m^2n^2=(mn)^2=\dim_{\mathbb C}M_{mn}(\mathbb C). \end{align*} A surjective linear map between finite-dimensional complex vector spaces of equal dimension is injective. Hence $\Phi_0$ is a bijective unital $*$-homomorphism. [/step]

[step:Pass from the algebraic tensor product to the minimal tensor product] By the definition of the minimal tensor norm, using the faithful standard representations of $M_m(\mathbb C)$ on $H_m$ and $M_n(\mathbb C)$ on $H_n$, the norm of $x\in M_m(\mathbb C)\odot M_n(\mathbb C)$ is \begin{align*} \|x\|_{\min}=\|(\pi_m\odot\pi_n)(x)\|_{\mathcal L(H_m\otimes H_n)}. \end{align*} Under the unitary identification $U:H_m\otimes H_n\to \mathbb C^{mn}$, the operator $(\pi_m\odot\pi_n)(x)$ is conjugate to $\Phi_0(x)$ for every $x\in M_m(\mathbb C)\odot M_n(\mathbb C)$. Thus \begin{align*} \|x\|_{\min}=\|U(\pi_m\odot\pi_n)(x)U^{-1}\|_{\mathcal L(\mathbb C^{mn})}=\|\Phi_0(x)\|_{M_{mn}(\mathbb C)}. \end{align*} So $\Phi_0$ is isometric for the minimal tensor norm and the usual operator norm on $M_{mn}(\mathbb C)$. Because $M_m(\mathbb C)\odot M_n(\mathbb C)$ is finite-dimensional, it is complete for the norm $\|\cdot\|_{\min}$. Hence its completion is the same [vector space](/page/Vector%20Space), now denoted $M_m(\mathbb C)\otimes_{\min}M_n(\mathbb C)$. Therefore $\Phi_0$ extends, with the same formula, to a unital isometric $*$-isomorphism \begin{align*} \Phi:M_m(\mathbb C)\otimes_{\min}M_n(\mathbb C)\to M_{mn}(\mathbb C). \end{align*} This is precisely the asserted map on matrix units, and the proof is complete. [/step]