Let $S=\{P_1,P_2,\dots,P_n\}$ be a set of Hilbert space projections $P_i=P_i^2=P_i^*$ that only commute when they are orthogonal or equal:
$$P_iP_j=P_jP_i\implies P_iP_j=0\text{ or }i=j.$$
Let $i,j,k$ be such that $P_iP_j\neq P_jP_i$ and $P_jP_k\neq P_kP_j$, then I can show that
$$P_iP_jP_k\neq 0.$$
That means that $P_iP_jP_k=0$ only when $P_iP_j=0$ or $P_jP_k=0$.
Is it possible to show this more generally? Or is there a counterexample? To formulate:
Let $T:=P_{i_1}P_{i_2}\cdots P_{i_k}$ be a product of projections
from $S$. Then $T=0$ only when some pair $P_{i_m}P_{i_{m+1}}=0$ are
orthogonal.True or False?
I can't seem to find any literature that might discuss such a problem.
Best Answer
Here is a counterexample: denote $$ A = \pmatrix{1 & 0\\0 & 0}, \quad B = \frac 12 \pmatrix{1&1\\1&1}, \quad C = \frac 15 \pmatrix{1&2\\2&4}, $$ and define the block-diagonal projections $$ P_1 = \pmatrix{A\\&A\\&&0}, \quad P_2 = \pmatrix{B\\ &0\\&&B}, \quad P_3 = \pmatrix{0 \\ & C\\ && C}. $$ Verify that for each pair $i \neq j$, $P_iP_j \neq P_j P_i$. Nevertheless, we have $P_1P_2P_3 = 0$.