Can a sum of idempotents vanish? Let $A$ be a finite dimensional $\mathbb C$-algebra. Let $e_1,\ldots,e_r\in A$ be nonzero idempotents (with $r>0$), i.e. $e_i^2=e_i$. My question is: Can it happen that $e_1+\cdots+e_r=0$? I can't think of a single example. 
Note: I do not require the $e_i$ to be central, primitive, or orthogonal.
 A: We can WLOG assume that the algebra $A$ is embedded in $\mathrm{M}_n\left(\mathbb C\right)$ for some $n\in\mathbb N$ (because the $A$-module $A$ is faithful and finite-dimensional, so that $A$ is embedded in $\mathrm{End}_{\mathbb C} A \cong \mathrm{M}_n\left(\mathbb C\right)$ for $n=\dim_{\mathbb C}A$). Then, $e_1, e_2, \dots, e_r$ are idempotent matrices, and have idempotent sum (because $0$ is idempotent). According to MathOverflow question #115067, any finite list of idempotent matrices over $\mathbb C$ (or any other field of characteristic $0$) having idempotent sum must be a list of orthogonal idempotents. Hence, your idempotents $e_1, e_2, \dots, e_r$ are orthogonal. Thus, $e_1\left(e_1+e_2+\cdots+e_r\right) = e_1e_1 + e_1e_2 + \cdots + e_1e_r = e_1 + 0 + \cdots + 0 = e_1$. Since $e_1+e_2+\cdots+e_r=0$, this rewrites as $e_1\cdot 0 = e_1$, so that $e_1 = 0$. This contradicts the assumption that $e_1, e_2, \dots, e_r$ are nonzero idempotents.
A: Here is a proof which does not make use of the hypothesis that $A$ is finite-dimensional. It works over any field of characteristic $0$. Since $A$ acts on itself by left multiplication, it suffices to answer this question for idempotent endomorphisms $e_1, e_2, ... e_r$ of some vector space $V$ (not necessarily finite-dimensional). Let $V_1, V_2, ... V_r$ be the images of $e_1, e_2, \dots, e_r$. If $V_1 \cap V_2$ is nonzero, pick direct sum decompositions
$$V_1 \cong V_1' \oplus (V_1 \cap V_2)$$
$$V_2 \cong V_2' \oplus (V_1 \cap V_2)$$
so that $e_1 = e_1' + e_{1 \cap 2}$ and $e_2 = e_2' + e_{1 \cap 2}$ where $e_i'$ is the projection onto $V_i'$ and $e_{1 \cap 2}$ is the projection onto $V_1 \cap V_2$. Now $e_1 + e_2 = e_1' + e_2' + 2 e_{1 \cap 2}$ where all three idempotents appearing in this sum are orthogonal. Next, consider the intersections $V_1' \cap V_3, V_2' \cap V_3, V_1 \cap V_2 \cap V_3$, and continue decomposing the idempotents by choosing direct sum decompositions in this manner. In the end we will have written $e_1 + e_2 + \cdots + e_r$ as a sum, with positive integer coefficients, of nonzero orthogonal idempotents, and over a field of characteristic zero such a sum is clearly nonzero (e.g. because we can pick a nonzero vector in the image of each idempotent, and their sum is not sent to zero). 
A: Only for zero idempotents. There are several articles on sum of idempotents. See, for example,
R. E. Hartwig, M. S. Putcha. When is a matrix a sum of idempotents?  Linear and Multilinear Algebra, 26 (1990) 279--286,
and for infinite dimensional case -
C. Pearcy  and D. M. Topping.   Sums of small numbers of idempotents. Michigan Math. J. 14, (1967), 453--465.
By the way, finite dimensional restriction is essential, since every bounded operator is a sum of 5 idempotents.
