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Would like to have some guidance. $P$ is projection matrix on $U$ and $0\notin v\notin \mathbb{R}^2$

I need to show that if $v$ is element of $U$ than $v$ is Eigenvector of $P$ with Eigenvalue 1.

I know that for projection matrix Eigenvalue is $1$ or $0$... but why in this case only $1$?

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up vote 4 down vote accepted

Well here I think that you mean that if v in U than v is an eigenvector of P (you said A) with eigenvalue 1. I think all you need here is the fact that P is (By definition projection ONTO U), so what happens to a v in U under the projection to U by P?... it projects it to itself. Ie if v is not 0 and v in U, Pv = v!

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