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Let H be a Hilbert space and M be a closed linear subspace of H.

Than for each $ x\in H$ can be uniquely expressed as $x=x_1+x_2$ where $x_1 \in M$ and $x_2\in M^\perp$.

The operator $p:H\rightarrow M$ defined by $p(x)=x_1\, \forall x\in H$ is called the projection operator of H on M.

How to prove that
1.$p$ is onto,
2.$\ker p=M^\perp$ and

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up vote 1 down vote accepted
  1. To show onto, you need to show that $\forall m \in M$, there exists an $x\in H$ such that $p(x) = m$. Given $m$, what would you try for $x$?

  2. $x \in \ker p$ means that $p(x) = 0$. Show that $M^{\perp } \subset \ker p \subset M^{\perp}$.

  3. $p^2 (x_1 + x_2) = p(x_1) = x_1$.

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