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My question is:

Let $H$ be a Hilbert space and $T \in B(H)$. Prove that $T$ is a projection if and only if $T$ is the identity on the orthogonal complement of its kernel.

Thanks

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closed as off-topic by Jack, Fabian, Davide Giraudo, Zachary Selk, RecklessReckoner Dec 26 '15 at 23:45

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Probably orthogonal projection. – abatkai Jun 3 '12 at 7:13
    
Do you need some other property, like self-adjoint? – copper.hat Jun 3 '12 at 7:25
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I assume "$T$ is a projection" is defined to mean "$T = T^*$ and $T^2 = T$." In this case, here is a hint (mainly for the "if" direction, which is the harder one): for any $A$ in $\mathcal{B}(H)$, the orthogonal complement of the kernel of $A$ is the closure of the range of $A^*$. – leslie townes Jun 3 '12 at 7:26
    
I know what you say. – Spring Xiao Jun 3 '12 at 8:11
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@leslietownes: If $T^*=T$ and $T^2=T$, then $T$ is an orthogonal projection (which is indeed what the question should specify). – robjohn Jun 3 '12 at 9:14

If "projection" is changed to "orthogonal projection", then here are some hints.

Hint 1: $\ker(T)^\perp=\{u\in H:Tx=0\Rightarrow \langle u,x\rangle=0\}$

Hint 2: Any $v\in H$ can be written as $v=u+k$ where $u\in\ker(T)^\perp$ and $k\in\ker(T)$.

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