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Show that if $P$ and $Q$ are two orthogonal projections with orthogonal ranges, then $P+Q$ is also an orthogonal projection.

First I need to show $(P+Q)^\ast = P+Q$. I am thinking that since \begin{align*} ((P+Q)^\ast f , g) & = (f,(P+Q)g) \\ & = (f,Pg) + (f,Qg) \\ & = (P^\ast f,g) + (Q^\ast f,g) \\ & = (Pf,g) + (Qf,g) \\ & = ((P+Q)f,g), \end{align*} we get $(P+Q)^\ast=P+Q$.

I am not sure if what I am thinking is right since I assumed that $(P+Q)f=Pf+Qf$ is true for any bounded linear operator $P$, $Q$.

For $(P+Q)^2=P+Q$, I use $$(P+Q)^2= P^2 + Q^2 + PQ +QP,$$ but I cant show $PQ=0$ and $QP=0$.

Anyone can help me? Thanks.

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    Looks good so far: Note that you didn't use that the ranges are orthogonal. This means that $\langle P f, Qg \rangle = 0$, among other things... Also: projections are linear.2012-08-06
  • 0
    +1 for showing a lot of your own thoughts. To get nicer formulas, you can enclose the formulas you wrote between dollar signs `$(P+Q)^\ast = P+Q$` gives $(P+Q)^\ast = P+Q$, for example. Single dollar sings for formulas in a paragraph, double dollar signs for displayed equations. Keep up the good work!2012-08-06

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