3
$\begingroup$

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.

  • 0
    @Rasmus the problem doesn't say something about the orthogonality of the projections2011-06-24

1 Answers 1

3

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.

  • 0
    Oh I see, it makes sense now. :) Thanks2011-06-24