# Show that $P_i$ and $\sum_i P_i$ being idempotent implies $P_i P_j=\delta_{ij}$

Let $$X$$ be a finite dimensional real linear space, or more generally a finite dimensional vector space over a field of characteristic $$0$$. Let $$(P_i)_{i=1}^n$$ be a finite sequence of linear mappings $$P_i :X\rightarrow X$$ such that

1. $$P_i^2=P_i$$ for $$i=1,...,n$$,
2. $$(P_1+...+P_n)^2=P_1+...+P_n$$.

I wish to show that $$P_i\circ P_j=0 \textrm{ for } i \neq j.$$

I know how to prove it only for $$n=2$$: if

$$(P_1+P_2)^2=P_1+P_2$$ then

$$P_1P_2+P_2P_1=0. \tag{\ast}$$

By multiplying both sides this equality by $$P_1$$ from left and by $$P_1$$ by right and obtain two equalities: $$P_1 P_2+P_1P_2P_1=0$$ and $$P_1P_2P_1+P_2P_1=0$$. By subtracting:

$$P_1P_2-P_2P_1=0. \tag{\ast\ast}$$

From $$(\ast)$$, $$({\ast}\ast)$$, we get $$P_1P_2=0$$, $$P_2P_1=0$$.

First, note that the condition that $$X$$ has finite dimension seems actually important (see linked question in comment below).

An important property of projectors in finite dimensional spaces (over fields with characteristic $$0$$) is that their trace coincides with their rank (indeed, since $$0$$ and $$1$$ are the only eigenvalues, both the rank and the trace count the number of ones).

Let $$p=p_1+p_2+\ldots +p_n$$, $$K={\sf Ker}(p),A={\sf Im}(p)$$ and $$A_i={\sf Im}(p_i)$$. By the remark made just above,

$${\sf dim}(A)=\sum_{k=1}^n {\sf dim}(A_k), \ A \subseteq \sum_{k=1}^n A_k \tag{1}$$

The two facts above imply that $$A$$ is the direct sum of the $$A_k$$. Since $$p$$ is a projector, $$X=K \oplus A$$, and hence

$$X=K \oplus \bigoplus_{k=1}^n A_k \tag{2}$$

For $$k\in K$$, one has $$0=pk=\sum_{j=1}^n p_jk$$. By the unicity in decomposition (2) above, we deduce that

$$p_j \ \text{is zero on} \ K \ (1\leq j\leq n) \tag{3}$$

Now, let $$q_{ij}$$ be the unique endomorphism of $$X$$ that coincides with $$p_i$$ on $$A_j$$, and is zero on $$K$$ and $$\bigoplus_{k\leq j}A_k$$. By contruction, those $$n^2$$ endomorphisms $$q_{ij} (1\leq i,j \leq n)$$ are linearly independent, and we have $$p_i=\displaystyle\sum_{j}q_{ij}$$ for every $$i$$, so

$$p=\sum_{i,j} q_{ij} \tag{4}$$

On the other hand, since $$p_i$$ is a projector it is the identity on its image $$A_i$$, $$q_{ii}$$ must be the identity on $$A_i$$, so

$$p=\sum_{i} q_{ii} \tag{5}$$

Combining (4) with (5), we see that $$\sum_{i\neq j}q_{ij}=0$$. By the linear independence of the $$q_{ij}$$, we deduce $$q_{ij}=0$$ for any $$i\neq j$$. So each $$p_i$$ reduces to $$q_{ii}$$, and $$p_i$$ is the projector onto $$A_i$$ according to $$K\oplus \bigoplus_{k\leq j}A_k$$. The claimed property is now clear.

• Is there an infinite-dimensional counterexample for n=3? – zyx May 2 '16 at 17:38
• @zyx It is very likely that there is. I've just asked a question about that here – Ewan Delanoy May 4 '16 at 9:25