The matrix-ring Morita pair with explicit tensor inverses

(The four subspaces.) Since $b{E}_{11}$ is the matrix whose first column is the first column of $b$ and whose other columns vanish, $Be=\{{\sum }_i{a}_i{E}_{i1}:a_i\in k\}$; dually $eB=\{{\sum }_j{c}_j{E}_{1j}:c_j\in k\}$, and $eBe=k{E}_{11}$ with $e=E_{11}$ as identity, so $A=k{E}_{11}\cong k$ as a field, the isomorphism being $\lambda \mapsto \lambda e$. The products land where claimed because $E_{1j}{E}_{i1}={\delta }_{ji}{E}_{11}$ and $E_{i1}{E}_{1j}=E_{ij}$ by [F1].

(The bimodule structures.) Left multiplication by $B$ and right multiplication by $A\subseteq B$ make $M=Be$ a $(B,A)$-bimodule: both actions are $k$-linear, and associativity of matrix multiplication gives $b(ma)=(bm)a$ for $b\in B$, $m\in Be$, $a\in A$, with $A=ke$ acting by scalar multiplication by [F2]. Symmetrically $N=eB$ is an $(A,B)$-bimodule.

(${\mu }_N$.) The pairing $(x,y)\mapsto xy$ from $eB\times Be$ to $eBe$ is $k$-bilinear and $B$-balanced: $((xb)y)=x(by)$ by associativity for $b\in B$; by [F3] it descends to a homomorphism ${\mu }_N$ with ${\mu }_N(x\otimes y)=xy$, which is a map of $(A,A)$-bimodules because both the product and the tensor actions are induced from the two factors. The map $eBe\to eB{\otimes }_{B}Be$, $a\mapsto e\otimes a$, is well defined, and for $x\in eB$, $y\in Be$ one has $x=ex$ and hence $x\otimes y=(ex)\otimes y=e\otimes (xy)$ by $B$-balance, so the two composites are the identities: ${\mu }_N(e\otimes a)=a$ for $a\in eBe$ and $e\otimes xy=x\otimes y$. Hence ${\mu }_N$ is an isomorphism of $(A,A)$-bimodules.

(${\mu }_M$.) The pairing $(m,n)\mapsto mn$ from $Be\times eB$ to $B$ is $k$-bilinear and balanced over $A=k{E}_{11}$: $(m{E}_{11})n=m(E_{11}n)$ is a case of associativity, and scalar balancing holds because $A=k$ acts as scalars by [F2]. By [F3] it descends to a homomorphism ${\mu }_M$ with ${\mu }_M(m\otimes n)=mn$, a map of $(B,B)$-bimodules. Define the $k$-linear inverse on the basis $\{E_{ij}\}$ by $E_{ij}\mapsto E_{i1}\otimes E_{1j}$; this is well defined because the matrix units form a $k$-basis of $B$, and ${\mu }_M(E_{i1}\otimes E_{1j})=E_{i1}{E}_{1j}=E_{ij}$. Conversely, writing $m={\sum }_i{a}_i{E}_{i1}$ and $n={\sum }_j{c}_j{E}_{1j}$ one has $m\otimes n={\sum }_{i,j}{a}_i{c}_j{E}_{i1}\otimes E_{1j}$ and $mn={\sum }_{i,j}{a}_i{c}_j{E}_{ij}$, so the two composites are the identities on a spanning set and hence everywhere. Thus ${\mu }_M$ is an isomorphism of $(B,B)$-bimodules.

(Conclusion.) Step 3.1 gives the bimodule isomorphism $N{\otimes }_{B}M\cong eBe=A$ and step 3.2 gives $M{\otimes }_{A}N\cong B$, so by [F4] the rings $A\cong k$ and $B=M_n(k)$ are Morita equivalent with inverse pair of bimodules $(M,N)$; the explicit inverses are $a\mapsto e\otimes a$ and $E_{ij}\mapsto E_{i1}\otimes E_{1j}$ extended $k$-linearly, and the only elements used are the fixed matrix units, so no choice is used.