little question about linear operators Let H be a complex Hilbert Space. Let $P \in L(H)$ be an idempotent operator ($P^{2} = P$). Also, let $\parallel P\parallel = 1$. I want to prove that $P$ is an orthogonal operator.  I defined $M = Im(P)$ and i proved that $M$ is a closed subset of H such that if $x \in M$, then $P(x)=x$. However, I couldn't prove that $x - P(x) \perp P(x)$. I would be able to do this, if i can prove that $P$ in this conditions is self-adjoint. However, apparently im stupid enough not to be able to do this.
Can you help me ? Thanks :)
 A: Assume $\|P\|\leq 1$. 
As you said, you can show that $P^*=P$ by proving that $H=\mbox{Ker} (I-P)\oplus \mbox{Ker} P$ is an orthogonal decomposition. Indeed, it is then stable under $P^*$, so $P^*=P$ is obvious on each summand.
So take $x$ in $\mbox{Im} P=\mbox{Ker}(I-P)$ and $y$ in $\mbox{Ker} P$. We need to show that $(x,y)=0$. 
Note that for every $t\in\mathbb{R}$, we have
$$
t^2\|x\|^2=\|P(tx+y)\|^2\leq \|tx+y\|^2=t^2\|x\|^2+2t\mbox{Re}(x,y)+\|y\|^2
$$
whence
$$
-2t\mbox{Re}(x,y)\leq \|y\|^2\qquad\forall t\in \mathbb{R}.
$$
It is clear that this entails $\mbox{Re}(x,y)=0$. Hence, applying this for $x$ and $e^{i\theta}y$, we get
$$
\mbox{Re}(x,e^{i\theta}y)=0\qquad \forall \theta\in\mathbb{R}.
$$
For the right choice of $\theta$, it follows that $(x,y)=0$. Precisely, if $(x,y)\neq 0$ and if the inner product is linear in the second variable, take 
$$
e^{i\theta}=\frac{\overline{(x,y)}}{|(x,y)|}.
$$
Note: it follows that a nonzero idempotent ($P^2=P$) is a projection ($P^2=P=P^*$) if and only if $\|P\|=1$. We have just shown $\Leftarrow$. The direction $\Rightarrow$ is clear from the orthogonal decomposition of $H$ above. Note also that $\|P\|\geq 1$ for any nonzero idempotent.
A: Here is a more algebraic and direct proof, using only operators and no vectors. 
From $\|P\|=1$ we get $P^*P\leq I$, so $I-P^*P\geq0$. Also, $P^*P=P^*P^*PP$, so $P^*(I-P^*P)P=0$. Then
$$
0=P^*(I-P^*P)P=((I-P^*P)^{1/2}P)^*(I-P^*P)^{1/2}P,
$$
which implies $(I-P^*P)^{1/2}P=0$. Multiplying by $(I-P^*P)^{1/2}$ on the left, we get
$$
(I-P^*P)P=0,
$$
which we can write as $P-P^*P^2=0$. So $P=P^*P^2=P^*P\geq0$. This shows that $P\geq0$ and that $P$ is selfadjoint. 
