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Please explain the specified phrase in more details:

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Ehsan zarei
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  • You asked a question before in similar style as you've done here. We cannot help you if you do not explain what you do not understand which you seem not to either understand or care about. – Cameron Williams Jun 13 '15 at 18:45

1 Answers1

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If $v \in M^\bot$, then $Pv=0$, since $P$ is the orthogonal projection on $M$.

Further, Proposition (3.1) shows that $M^\bot$ is invariant under each $\pi(x)$. Hence, $\pi(x)v \in M^\bot$. With the same argument as above, we get $P \pi(x)v=0$.

PhoemueX
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