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Let $p$ be a projection such that the projection $(I-p)$ is invertible. Does $(I-p)=(I+p)$ hold?

I proved it this way:

$$(I-p)^2=(I-p)=(I^2-p^2)=(I-p)(I+p)$$

Hence,

$$(I-p)^{-1}(I-p)^2= (I+p).$$

This is equivalent to

$$(I-p)=(I+p).$$

I used properties of projections $p^2=p$ and $(I-p)^2=(I-p)$.

  • If $I - p = I + p$ then $-p = p$ and by comparing each entry in $p$ we seethat $p = 0$ so $p$ is one of the two trivial projections (the other one being the "identity"). – William M. Jun 04 '21 at 15:29

1 Answers1

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Your proof is correct, but the result isn't very interesting. In fact, the only invertible projection $P'$ is $P' = I$ since $P' = (P')^{-1}(P')^2 = (P')^{-1}P' = I$. Thus $I - P$ invertible means $I - P = I$, i.e. $P = 0$.

Paul Frost
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