projection generated by intersection of two projection

Question

let $H$ be a Hilbert space and $P,Q$ be projections on $H$. suppose $P,Q$ do not commute. $P\wedge Q$ is a projection on $PH\cap QH$. I want to calculate $P\wedge Q$ but I can not. Please help me. Thanks in advance.

Answer 1

I don't think a formula can be given in finite terms, but we can construct $P\land Q$ as a strong limit as follows: Let $R_n = (PQP)^n$. Then the $R_n$ are a decreasing seqeuence of positive operators, hence converge strongly to some positive operator $R$. We will show $R = P \land Q$.

We have $$ (PQP)^nR = \lim_m R_{n}R_m = \lim_{m} R_{n+m} = R $$ and hence with $n \to\infty$, $R^2 = R$. So $R$ is a projection. Moreover $$ (PQP)^nQR = \lim (PQP)^nQ(PQP)^m = \lim (PQP)^{n+m+1} = R $$ hene $RQR = R$, giving $R \le Q$. As $$ P(PQP)^n = (PQP)^n \to R $$ we have $PR = R$, therefore $R \le P$. If finally $T$ is any projection with $T \le P, Q$, we have $TP = T$, $TQ = T$, hence $TR_n = T$, giving $TR = R$, that is $T \le R$. So $R = P \land Q$.