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Let $H$ be a Hilbert space and $P_1,P_2$ orthogonal projections on the closed spaces $M_1$ and $M_2$.

Let $\langle P_1x,x\rangle$ $\leq$ $\langle P_2x,x\rangle$ for all $x\in H$. Show that $P_1P_2=P_2P_1=P_1$

I am not able to get a connection between the assumption and the statement which I want to show..

My idea is to show that $\langle P_1P_2x,x\rangle=\langle P_2P_1x,x\rangle$ which would show that $P_1P_2=P_2P_1$

Pp..
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Epsilondelta
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  • Thanks again for editing. I need more practice with latex :) – Epsilondelta Feb 07 '15 at 16:45
  • I never use the angles for scalar products. They are a pain to type. And grew up with books that didn't use them. But it is true that parentheses can be confused with tuples. – Pp.. Feb 07 '15 at 16:51

1 Answers1

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When $P$ is orthogonal, $\langle Px,x\rangle=\langle Px,Px\rangle=||Px||^2$. With your hypothesys this gives that $\ker(P_2)\subseteq\ker(P_1)$, thus $M_1\subseteq M_2$. From this, the conclusion follows immediately.

Theo
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