An element $e$ in a K-algebra $A$ is called an idempotent if $e^2 = e$. The idempotents $e_1, e_2$ ∈ $A$ are called orthogonal if $e_1e_2 = e_2e_1 = 0$. The idempotent $e$ is said to be primitive if $e$ cannot be written as a sum $e = e_1 + e_2$, where $e_1$ and $e_2$ are nonzero orthogonal idempotents of $A$.
If $e$ is a central idempotent, then so is $1−e$, and hence $eA$ and $(1−e)A$ are two-sided ideals and they are easily shown to be K-algebras with identity elements $e$ ∈ $eA$ and $1 − e$ ∈ $(1 − e)A$, respectively. In this case the decomposition $A_A = eA⊕(1−e)A$ is a direct product decomposition of the algebra $A$.
Because the algebra $A$ is finite dimensional, the module $A_A$ admits a direct sum decomposition $A_A = P1 ⊕ · · · ⊕ Pn$, where $P_1, . . ., P_n$ are indecomposable right ideals of $A$. $P_1 = e_1A, . . . , P_n = e_nA$, where $e_1, . . . , e_n$ are primitive pairwise orthogonal idempotents of $A$ such that $1 = e_1 + · · · + e_n$. Conversely, every set of idempotents with the preceding properties induces a decomposition $A_A = P_1 ⊕ · · · ⊕ P_n$ with indecomposable right ideals $P_1 = e_1A, . . . , P_n = e_nA$.
Such a decomposition is called an indecomposable decomposition of $A$ and such a set ${e_1, · · · , e_n}$ is called a complete set of primitive orthogonal idempotents of $A$.
My question is: Why can I do the descomposition $A=\bigoplus e_iAe_j$, $i,j \in \left \{ 1,...,n \right \}$? Is it for use ideals of the form $e_iA, Ae_j$?
As an example, $\hom(M\oplus N, M\oplus N)\cong \begin{bmatrix}\hom(M,M)&\hom(N,M)\\ \hom(M,N)&\hom(N,N)\end{bmatrix}$. Both sides are rings, and they are isomorphic as rings. Notice that the multiplication is matrix multiplication on the right hand side though.
You can see how to split a map on the left into pieces by restricting the domain to pieces of the sum, and projecting to the factors in the image.
The same can be done for any finite direct sum of modules, where the entries are $\hom (M_i, M_j)$ in a matrix. In particular,
You can show that for any idempotents $e$ and $f$, $\hom_R(eR, fR)\cong fRe$ as abelian groups. Putting these things together, you can see where the isomorphism you’re talking about comes from.