TITLE: Does idempotent completion commute with direct sum completion?
QUESTION [5 upvotes]: Definitions: For a pre-additive category $\mathcal{C}$ I denote its idempotent completion by $\overline{\mathcal{C}}^p$. The objects of $\overline{\mathcal{C}}^p$ are pairs $(X,p)$, where $X \in \mathcal{C}$ and $p$ is an idempotent on $X$. A morphism $f:(X,p) \to (Y,q)$ is a morphism $f:x \to Y$ of $\mathcal{C}$ such that $fp = f$ and $qf=f$. The embedding $\mathcal{C} \to \overline{\mathcal{C}}^p$, $X \mapsto (X,1_X)$ satisfies the (weak) universal property of a completion.
Similarly, the direct sum completion $\overline{\mathcal{C}}^{\oplus}$ has as objects finite lists of objects of $\mathcal{C}$. A morphism $f: (X_1,\ldots, X_n) \to (Y_1, \ldots, Y_m)$ is a matrix $(f_{ij})$ of morphisms $f_{ij}:X_i \to X_j$. It is the universal additive category containing $\mathcal{C}$ (via the embedding $X \mapsto (X,1_X)$).
Question: Is there always a canonical equivalence $\overline{\mathcal{C}}^{p \oplus} \simeq \overline{\mathcal{C}}^{\oplus p}$. If not, is there a simple counterexample?
Additional info: The statement is equivalent to $\overline{\mathcal{C}}^{\oplus}$ being idempotent complete for any idempotent complete category $\mathcal{C}$ and $\overline{\mathcal{C}}^{p}$ being additive for any additive category $\mathcal{C}$. The latter is straightforward, giving rise to the embedding
$$
\overline{\mathcal{C}}^{p\oplus} \to \overline{\mathcal{C}}^{\oplus p} \\
((X_1, p_1),\ldots, (X_n, p_n)) \mapsto ((X_1,\ldots, X_n),(p_1 \oplus \ldots \oplus p_n))
$$
where the direct sums of morphisms is taking block matrices. ("A direct sum of subobjects is a subobject of the direct sum.") But I can't see how the other direction would follow in general.
Possible counter example: For a non semi-simple algebra $A$ let $\mathcal{A}$ be the 1-object subcategory of $A-\operatorname{Mod}$ that only contains the regular representation. Then $\overline{\mathcal{A}}^{\oplus p}$ is $\operatorname{proj}(A)$ but what is $\overline{\mathcal{A}}^{p\oplus}$? If $A$ is semi-simple then every $A$-module is projective so this is not a counter example in that case.
Origin of the question: It is claimed in passing in this paper on page 6 at the end of the first paragraph. So probably I'm just blind for not seeing it. Any help is appreciated.

REPLY [6 votes]: Indeed, your proposed counterexample works: $\overline{\mathcal{A}}^{p\oplus}$ will be the category of $A$-modules that are finite direct sums of direct summands of $A$.  In general, this will not include all the finitely generated projective modules.  For instance, if $A$ is a Dedekind domain that is not a PID, then it has no nontrivial direct summands so the idempotent completion of $\mathcal{A}$ just adds a zero object so $\overline{\mathcal{A}}^{p\oplus}$ is just the finitely generated free modules, but not every finitely generated projective module is free.