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.

  • 1
    $\begingroup$ Is there an infinite-dimensional counterexample for n=3? $\endgroup$ – zyx May 2 '16 at 17:38
  • $\begingroup$ @zyx It is very likely that there is. I've just asked a question about that here $\endgroup$ – Ewan Delanoy May 4 '16 at 9:25
  • $\begingroup$ @zyx Yes, in the paper Stampfli, Joseph G. Sums of projections. Duke Math. J. 31 (1964), no. 3, 455--461 an example with four projections is constructed that sum up to I but that are not disjoint. $\endgroup$ – Andrei Kh Jan 8 at 22:37

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