[proofplan]
We first construct the canonical map by using the lexicographic bijection between pairs of indices $(i,a)\in\{1,\dots,m\}\times\{1,\dots,n\}$ and single indices in $\{1,\dots,mn\}$. This turns an $m\times m$ matrix whose entries are $n\times n$ matrices over $A$ into one $mn\times mn$ matrix over $A$. We verify directly that this reindexing preserves multiplication and involution, then use [uniqueness of the C*-norm](/theorems/8546) to conclude that it is a C*-algebra *-isomorphism. Finally, Murray-von Neumann equivalence is preserved because tensoring the implementing partial isometry with $1_k$ preserves the two defining equations.
[/proofplan]
[step:Reindex pairs of matrix indices into a single matrix index]
For $m,n\in\mathbb N$, define the bijection
\begin{align*}
\theta_{m,n}:\{1,\dots,m\}\times\{1,\dots,n\}\to\{1,\dots,mn\},\qquad \theta_{m,n}(i,a)=(i-1)n+a.
\end{align*}
For $X\in M_m(M_n(A))$, write $X_{ij}\in M_n(A)$ for its $(i,j)$-entry and write $(X_{ij})_{ab}\in A$ for the $(a,b)$-entry of $X_{ij}$. Define
\begin{align*}
\Phi_{m,n}:M_m(M_n(A))\to M_{mn}(A)
\end{align*}
by the entry formula
\begin{align*}
(\Phi_{m,n}(X))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}=(X_{ij})_{ab}
\end{align*}
for all $i,j\in\{1,\dots,m\}$ and all $a,b\in\{1,\dots,n\}$.
Since $\theta_{m,n}$ is a bijection, every entry of a matrix in $M_{mn}(A)$ has a unique pair of row and column indices of the form $\theta_{m,n}(i,a)$ and $\theta_{m,n}(j,b)$. Therefore the displayed formula defines a bijective [linear map](/page/Linear%20Map), with inverse obtained by the same reindexing in the opposite direction.
[/step]
[step:Check that the reindexing preserves multiplication and involution]
Let $X,Y\in M_m(M_n(A))$. Fix $i,l\in\{1,\dots,m\}$ and $a,d\in\{1,\dots,n\}$. The $(\theta_{m,n}(i,a),\theta_{m,n}(l,d))$-entry of $\Phi_{m,n}(X)\Phi_{m,n}(Y)$ is
\begin{align*}
\sum_{j=1}^{m}\sum_{b=1}^{n}(\Phi_{m,n}(X))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}(\Phi_{m,n}(Y))_{\theta_{m,n}(j,b),\theta_{m,n}(l,d)}.
\end{align*}
By the definition of $\Phi_{m,n}$, this equals
\begin{align*}
\sum_{j=1}^{m}\sum_{b=1}^{n}(X_{ij})_{ab}(Y_{jl})_{bd}.
\end{align*}
For each fixed $j$, the inner sum is the $(a,d)$-entry of the product $X_{ij}Y_{jl}$ in $M_n(A)$. Hence the whole expression is
\begin{align*}
\left(\sum_{j=1}^{m}X_{ij}Y_{jl}\right)_{ad}.
\end{align*}
The matrix product in $M_m(M_n(A))$ gives
\begin{align*}
(XY)_{il}=\sum_{j=1}^{m}X_{ij}Y_{jl},
\end{align*}
so
\begin{align*}
(\Phi_{m,n}(X)\Phi_{m,n}(Y))_{\theta_{m,n}(i,a),\theta_{m,n}(l,d)}=(\Phi_{m,n}(XY))_{\theta_{m,n}(i,a),\theta_{m,n}(l,d)}.
\end{align*}
Since this holds for all entries, $\Phi_{m,n}(XY)=\Phi_{m,n}(X)\Phi_{m,n}(Y)$.
For the involution, fix $i,j\in\{1,\dots,m\}$ and $a,b\in\{1,\dots,n\}$. Then
\begin{align*}
(\Phi_{m,n}(X^*))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}=((X^*)_{ij})_{ab}.
\end{align*}
The involution in $M_m(M_n(A))$ gives $(X^*)_{ij}=X_{ji}^*$, and the involution in $M_n(A)$ gives $(X_{ji}^*)_{ab}=((X_{ji})_{ba})^*$. Therefore
\begin{align*}
(\Phi_{m,n}(X^*))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}=((X_{ji})_{ba})^*.
\end{align*}
By the definition of the involution in $M_{mn}(A)$, this is exactly
\begin{align*}
(\Phi_{m,n}(X)^*)_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}.
\end{align*}
Thus $\Phi_{m,n}(X^*)=\Phi_{m,n}(X)^*$.
[guided]
The point of this step is to verify that the reindexing is not only a relabelling of entries, but a relabelling compatible with the algebra structure. Let $X,Y\in M_m(M_n(A))$. The product $XY$ is computed first at the outer $m\times m$ level, and each outer product is then computed inside $M_n(A)$. Thus
\begin{align*}
(XY)_{il}=\sum_{j=1}^{m}X_{ij}Y_{jl}.
\end{align*}
Taking the $(a,d)$-entry inside $M_n(A)$ gives
\begin{align*}
((XY)_{il})_{ad}=\sum_{j=1}^{m}\sum_{b=1}^{n}(X_{ij})_{ab}(Y_{jl})_{bd}.
\end{align*}
Now compute the product after reindexing. The row index corresponding to $(i,a)$ is $\theta_{m,n}(i,a)$, and the column index corresponding to $(l,d)$ is $\theta_{m,n}(l,d)$. Matrix multiplication in $M_{mn}(A)$ gives
\begin{align*}
(\Phi_{m,n}(X)\Phi_{m,n}(Y))_{\theta_{m,n}(i,a),\theta_{m,n}(l,d)}=\sum_{j=1}^{m}\sum_{b=1}^{n}(\Phi_{m,n}(X))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}(\Phi_{m,n}(Y))_{\theta_{m,n}(j,b),\theta_{m,n}(l,d)}.
\end{align*}
Substituting the definition of $\Phi_{m,n}$ turns this into
\begin{align*}
\sum_{j=1}^{m}\sum_{b=1}^{n}(X_{ij})_{ab}(Y_{jl})_{bd}.
\end{align*}
This is the same expression as $((XY)_{il})_{ad}$, and therefore it is the entry of $\Phi_{m,n}(XY)$ in the same position. Since matrices over $A$ are equal exactly when all entries agree, $\Phi_{m,n}(XY)=\Phi_{m,n}(X)\Phi_{m,n}(Y)$.
For the involution, the only issue is the order reversal. In $M_m(M_n(A))$, the $(i,j)$-entry of $X^*$ is $X_{ji}^*$. Inside $M_n(A)$, the $(a,b)$-entry of $X_{ji}^*$ is $((X_{ji})_{ba})^*$. Hence
\begin{align*}
(\Phi_{m,n}(X^*))_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}=((X_{ji})_{ba})^*.
\end{align*}
But in $M_{mn}(A)$, the involution also swaps the two reindexed positions and takes the involution in $A$, so
\begin{align*}
(\Phi_{m,n}(X)^*)_{\theta_{m,n}(i,a),\theta_{m,n}(j,b)}=((X_{ji})_{ba})^*.
\end{align*}
Thus $\Phi_{m,n}(X^*)=\Phi_{m,n}(X)^*$. The same entrywise definition also preserves addition and scalar multiplication, so $\Phi_{m,n}$ is a bijective *-homomorphism at the algebraic level.
[/guided]
[/step]
[step:Use uniqueness of the C*-norm to identify the C*-algebras]
The preceding step shows that $\Phi_{m,n}$ is a bijective *-homomorphism between the involutive algebras $M_m(M_n(A))$ and $M_{mn}(A)$. The norm on $M_m(M_n(A))$ and the pullback norm
\begin{align*}
X\mapsto \|\Phi_{m,n}(X)\|_{M_{mn}(A)}
\end{align*}
both make the same involutive algebra $M_m(M_n(A))$ into a C*-algebra. By [citetheorem:8546], these two C*-norms agree. Therefore $\Phi_{m,n}$ is isometric. Since it is also bijective and preserves multiplication and involution, it is a C*-algebra *-isomorphism.
[/step]
[step:Tensor an implementing partial isometry with the identity matrix]
Assume that $p,q\in M_n(A)$ are projections and that $p$ is Murray-von Neumann equivalent to $q$. By definition, there exists an element $v\in M_n(A)$ such that
\begin{align*}
v^*v=p
\end{align*}
and
\begin{align*}
vv^*=q.
\end{align*}
Fix $k\in\mathbb N$, and let $1_k\in M_k(\mathbb C)$ denote the identity matrix. Define
\begin{align*}
w:=v\otimes 1_k\in M_n(A)\otimes M_k(\mathbb C)\cong M_{nk}(A).
\end{align*}
Using the product and involution in the [tensor product](/page/Tensor%20Product) C*-algebra, we compute
\begin{align*}
w^*w=(v\otimes 1_k)^*(v\otimes 1_k)=(v^*\otimes 1_k)(v\otimes 1_k)=v^*v\otimes 1_k=p\otimes 1_k.
\end{align*}
Similarly,
\begin{align*}
ww^*=(v\otimes 1_k)(v\otimes 1_k)^*=(v\otimes 1_k)(v^*\otimes 1_k)=vv^*\otimes 1_k=q\otimes 1_k.
\end{align*}
Thus $w$ implements Murray-von Neumann equivalence between $p\otimes 1_k$ and $q\otimes 1_k$ inside $M_n(A)\otimes M_k(\mathbb C)\cong M_{nk}(A)$. This proves the stability of Murray-von Neumann equivalence under finite diagonal matrix amplification.
[/step]
