I am reading an article and it is written there: If A is a k-linear category (possibly without direct sums) we can embed it in the additive category A × N, where a morphism (x,m) → (y, n) is an n × m matrix with entries in A(x, y) = HomA(x, y). Of course if A is additive then A ≈ A × N. Could anyone explain it? Why is it additive?



embedding a k-linear category in an additive category
It looks like if this is a wrong construction of the additivization of a $k$-linear category. Either this, or I don't understand the notation. Let me just sketch how the additivization $\operatorname{Add}(\mathcal{A})$ of a $k$-linear category $\mathcal{A}$ can be constructed:


Objects in $\operatorname{Add}(\mathcal{A})$ are finite sequences $(X\_1,\dots, X\_n)$ of objects in $\mathcal{A}$, $n\geq 0$. Morphism $k$-modules are as follows:


$$\operatorname{Add}(\mathcal{A})((X\_1,\dots, X\_n),(Y\_1,\dots, Y\_m))=
\bigoplus\_{i=1}^n\bigoplus\_{j=1}^m\mathcal{A}(X\_i,Y\_j).
$$
In particular, a morphism $\varphi\colon (X\_1,\dots, X\_n)\rightarrow(Y\_1,\dots, Y\_m)$ is a matrix $\varphi=(\varphi\_{ij})$ formed by morphisms $\varphi\_{ij}\colon X\_i\rightarrow Y\_j$. Composition is defined by matrix multiplication and composition in $\mathcal{A}$. Direct sums can be calculated as
$$(X\_1,\dots, X\_n)\oplus(Y\_1,\dots, Y\_m)=(X\_1,\dots, X\_n,Y\_1,\dots, Y\_m).$$
The empty sequence $()$ is a zero object.


The $k$-linear functor
$$
\mathcal{A}\longrightarrow\operatorname{Add}(\mathcal{A}),$$
$$X\mapsto(X),$$
is fully faithful and universal (initial) among $k$-linear functors from $\mathcal{A}$ to $k$-linear additive categories.


If $\mathcal{A}$ is additive, a choice of direct sums defines an inverse equivalence,
$$
\operatorname{Add}(\mathcal{A})\longrightarrow \mathcal{A},
$$
$$(X\_1,\dots, X\_n)\mapsto X\_1\oplus\cdots\oplus X\_n.$$


