6
$\begingroup$

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.

  • 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

1 Answers 1

3

To complete your proof we need the following observations.

If $\langle f,g\rangle=0$ for all $g\in H$, then $f=0$. Indeed, take $g=f$, then you get $\langle f,g\rangle=0$. By definition of inner product this implies $f=0$.

Since $\mathrm{Im}(P)\perp\mathrm{Im}(Q)$, then for all $f,g\in H$ we have $\langle Pf,Qg\rangle=0$. which is equivalent to $\langle Q^*Pf,g\rangle=0$ for all $f,g\in H$. Using observation from previous section we see that $Q^*P(f)=0$ for all $f\in H$, i.e. $Q^*P=0$. Since $Q^*=Q$ and $P^*=P$ we conclude $ QP=Q^*P=0 $ $ PQ=P^*Q^*=(QP)^*=0^*=0 $

In fact $R$ is an orthogonal projection iff $R=R^*=R^2$. In this case we can prove your result almost algebraically $ (P+Q)^*=P^*+Q^*=P+Q $ $ (P+Q)^2=P^2+PQ+QP+Q^2=P+0+0+Q=P+Q $ I said almost, because prove of $PQ=QP=0$ requires some machinery with elements of $H$ and its inner product.

  • 0
    @t.b. added, thanks for your comments and suggestions2012-08-06