# orthogonal projection relation

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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For (1), start with $(PQ)^2 = PQPQ$ and use the premise that $PQ=QP$ to simplify. For (2), it depends on whether $PQ,P,Q$ are orthogonal or not. – Christian Clason Nov 28 '12 at 15:06
Yes, PQ, P,Q are all orthogonal – Voyage Nov 28 '12 at 15:09
Then you can use that any vector $v$ can be decomposed as $v=v_1 + v_2$, where (for example) $v_1\in ker P$ and $v_2 \in (ker P)^\bot = ran(P)$. – Christian Clason Nov 28 '12 at 15:12
The first statement is false, even assuming that everything is orthogonal: take $P$ any orthogonal operator, and $Q$ to be the identity. Then $PQ=QP=P$ but $P$ need not be a projection. – Chris Eagle Nov 28 '12 at 15:15
You should edit your question to include this information. – Christian Clason Nov 28 '12 at 15:18

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 $\ker PQ\subset\ker P+\ker Q$.