# Necessary and sufficient conditions for a sum of idempotents to be idempotent

Given: a finite set of n by n idempotent complex matrices. If pairwise products are zero, it is trivial to show the sum is idempotent. The converse, while true, is not so trivial: If the sum is idempotent, all pairwise products are zero.

I have been told that the converse is well known among statisticians but I cannot find a reference. Do you know one?

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What does NASC mean? – Todd Trimble Dec 1 '12 at 13:45
Necessary And Sufficient Conditions, apparently. – Allen Knutson Dec 1 '12 at 16:47
Ah, thanks Allen. – Todd Trimble Dec 1 '12 at 17:29
I am NOT assuming matrices are self adjoint. I'm not looking for a proof, just a reference. – user29604 Dec 3 '12 at 3:06

The key point is that the image of the sum of the idempotents is necessarily the direct sum of the images of the individual idempotents.

Suppose that $E_1,\ldots,E_k$ are idempotent and $F = \sum E_i$ is idempotent. Let $R_i$ and $K_i$ be the image and kernel of $E_i$, respectively, and let $R$ be the image of $F$.

The trace of an idempotent equals its rank, so $$\dim R = \mathrm{tr}(F) = \sum \mathrm{tr}(E_i) = \sum \dim R_i.$$ Furthermore $R$ is a subspace of $\sum R_i$ and $\sum R_i$ has dimension at most $\sum \dim R_i$, with equality iff this sum is direct, so we actually have $$R = R_1\oplus \cdots \oplus R_k.$$ Consider a vector $v_1\in R_1$. By definition $$(v_1,0,\ldots,0) = Fv_1 = (E_1v_1,E_2v_1,\ldots,E_k v_1)$$ with respect to this decomposition. In other words, $R_1\subset K_i$ for all $i\neq 1$, or $E_i E_1 = 0$ for all $i\neq 1$. It follows similarly that $E_i E_j = 0$ for all $i\neq j$.

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