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Let $P$, $Q:E\rightarrow E$, be projections and $PQ=QP$, show that $N(P)+N(Q)=N(PQ)$, $N(P)$ stands for Kernel of $P$

As $P$, $Q$ are projections and $PQ=QP$ then $PQ$ is a projection, so $E= N(PQ)\oplus \text{Im}(PQ)$ and also $E=N(P)\oplus \text{Im}(P)=N(Q)\oplus \text{Im}(Q)$. It's easy to see $N(P)+N(Q)\subset N(PQ)$ , but don't know how to arrive to the other inclusion or how to put $E=N(P)+N(Q)\oplus \mathrm{Im}(PQ)$ in that case the answer will follow immediately.

I've tried this last idea by writing $2v=(v-Pv)+(v-Qv)+(Pv+Qv)$ because $(v-Pv)+(v-Qv)$ is in $N(PQ)$ but then I can't conclude that Pv+Qv is in $\mathrm{Im}(PQ)$.

Any hint would be appreciated, Thanks in advance.

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I think you should provide a definition for the nucleus of a projection. It's what's more widely known as the kernel, isn't it? –  Rasmus Jun 24 '11 at 6:54
    
Oh yes sorry you're right Nucleus is Kernel I'll edit it –  Ivan3.14 Jun 24 '11 at 6:59
    
Also, are you talking about orthogonal projections in Hilbert space? –  Rasmus Jun 24 '11 at 7:05
    
@Rasmus: I think that the relations $P^2=P$, $Q^2=Q$ and $QP=PQ$ suffice. The first two (idempotency) relations mean that that the two endomorphisms are projections, and the last was given (and also implies that the composite mapping is an idempotent). No need to assume an inner product, I think. –  Jyrki Lahtonen Jun 24 '11 at 7:14
    
@Rasmus the problem doesn't say something about the orthogonality of the projections –  Ivan3.14 Jun 24 '11 at 7:16

1 Answer 1

up vote 3 down vote accepted

If $PQv=0$, then $v=v-PQv=(v-Pv)+(Pv-PQv)$. Here clearly $v-Pv$ is in the kernel of $P$. As $PQ=QP$, the latter term really is $(Pv-PQv)=(Pv)-Q(Pv)$, and this is in the kernel of $Q$. We have written an arbitrary vector annihilated by $PQ$ as a sum of a vector from $\mathrm{Ker}(P)$ and a vector from $\mathrm{Ker}(Q)$. Therefore the missing inclusion is proven.

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Jyrki, could you please explain your last step, namely getting to $(v-Pv)+(Pv-PQv)$? Thanks –  InterestedGuest Jun 24 '11 at 7:30
    
@InterestedGuest: Did you mean how to get $$v=v+0-0=v+(Pv-Pv)-PQv=(v-Pv)+(Pv-PQv)$$ or what? I simply subtracted a zero vector ($=PQv$), added and subtracted $Pv$, and did a bit of regrouping. –  Jyrki Lahtonen Jun 24 '11 at 7:43
    
Oh I see, it makes sense now. :) Thanks –  InterestedGuest Jun 24 '11 at 7:51

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