Let $X$ be a Hilbert space and $A\subset X$ a closed subspace show that the orthognal projection $P:X\rightarrow A$ is linear.
Now I know that $x-P(x)\in A^{\perp}$. The lecture notes go on by saying that since
$x-P(x)\in A^{\perp}$, $\alpha x-\alpha P(x)\in A^{\perp}$ for $\alpha$ a scalar of the field. Since this is also the case for $\alpha x-P(\alpha x)$ they conclude that $\alpha P(x)=P(\alpha x)$. Now the last conclusion is what I don't quite understand.
By the same logic they proved that $P(x+y)=P(x)+P(y)$ which therefore I didn't really understand neither.
Best Answer
$\alpha x-\alpha Px \in A^{\perp}$ and $\alpha x-P(\alpha x )\in A^{\perp}$ together imply that $P(\alpha x )-\alpha Px \in A^{\perp}$ because $P(\alpha x )-\alpha Px=[\alpha x-\alpha Px] -[\alpha x-P(\alpha x )]$. Hence $P(\alpha x )-\alpha Px \in A^{\perp}$ and $P(\alpha x )-\alpha Px \in A$, so $P(\alpha x )-\alpha Px =0$. Similar argument for $P(x+y)$.