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I try to solve some exercises from Conway "Functional Analysis", and I have some problems with the following: $P,Q,PQ$ are orthogonal projections.

(1) $PQ$ is orthogonal projection iff $PQ=QP$

(2) $\ker(PQ)$=$\ker(P)+\ker(Q)$

I already proved that if $PQ$ is a projection then $PQ=QP$. For the other direction I need to show that $PQ=(PQ)^2$ (because then we know that $PQ$ is idempotent), but I have no idea how to prove it.

For (2) I have no idea.

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    You should edit your question to include this information.2012-11-28

1 Answers 1

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You are not saying it in the question, but you are assuming that both $P,Q$ are orthogonal projections (i.e. $P^*P=P$, $Q^*Q=Q$).

For (1), if $PQ=QP$, then $(PQ)^2=PQPQ=PPQQ=PQ$, and $(PQ)^*=QP=PQ$.

For (2), you are assuming (again without saying!) that $PQ$ is a projection. First, if $x\in\ker P+\ker Q$, then $x=y+z$ with $Py=0$, $Qz=0$. Then $ PQx=PQ(y+z)=PQy+PQz=QPy+PQz=0+0=0. $ So $\ker P+\ker Q\subset\ker PQ$. Conversely, if $x\in\ker PQ$, then $PQx=0$, so $Qx=(I-P)Qx$. Then $x=Qx+(I-Q)x=(I-P)Qx+(I-Q)x$, and $(I-P)Qx\in\ker P$, $(I-Q)x\in\ker Q$. So $\ker PQ\subset\ker P+\ker Q$.