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Given $p_1, ..., p_n$ $n$ projection operators on the vector space $E$ such that $\sum_{i=1}^n p_i$ is a projection operator.

How to show that $\forall i,j \text{ s.t. } i \neq j, p_i \circ p_j = 0$ ?

I showed this for the case $n=2$ but I can't extend it by induction.

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  • $\begingroup$ You work here, I supose, with some inner product, don't you? $\endgroup$
    – DonAntonio
    Commented May 27, 2016 at 9:46
  • $\begingroup$ @Joanpemo : I'm not sure what you mean but $\circ$ here is the composition law. $\endgroup$
    – servabat
    Commented May 27, 2016 at 14:07
  • $\begingroup$ Yes, I know that, thank you. I was asking because I began thinking of a probable proof using inner product, yet I'm not that sure already. $\endgroup$
    – DonAntonio
    Commented May 27, 2016 at 14:20
  • $\begingroup$ @Joanpemo : sorry I'm not familiar with the english vocabulary. No, $E$ is not an inner product space, but only a vector space. Altought, a proof, even assuming $E$ is an inner product space, would be very interesting, and I'd love to see it :) $\endgroup$
    – servabat
    Commented May 27, 2016 at 14:32

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