Hilbert Space: Orthogonal projection is linear

Question

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.

Answer 1

$\alpha x-\alpha Px \in A^{\perp}$ 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 $P(x+y)$.

Answer 2

It depends on your definition of orthogonal projections in a Hilbert space, whether the conclusion makes sense or not.

Normally, the above conclusion should be a direct consequence of the uniqueness of the projection, i.e. for $x \in X$, the projection $P_A(x)$ is the unique element of $A$ such that $x - P_A(x) \in A^\perp$. The existence and uniqueness of this element form the fundamental basis of Hilbert space theory.

Now, for $\alpha \in \mathbb{C}$ you have that $\alpha(P_A(x))$ and $P_A(\alpha x)$ fulfill the same condition, namely

$$ \alpha x - \alpha P_A(x) \in A^\perp \quad\text{and}\quad \alpha x - P_A(\alpha x) \in A^\perp. $$

By uniqueness, both have to coincide.