For an exercise, I need to investigate the orthogonal projection in more detail. I know that this has probably been discussed many times here, I am though interested in whether my explanations suffice and in particular in Tasks 3 and 4. Concretely, the exercise states :

Let $\mathcal{M}$ be a closed subspace of a Hilbert space $\mathcal{H}$ and let

$\mathcal{M}^{\bot} = \{x \in \mathcal{H} : \langle x, y \rangle\_{\mathcal{H}} = 0 , \forall y \in \mathcal{M} \}$

be the corresponding orthogonal space, i.e. the space of all vectors $x \in \mathcal{H}$ which are perpendicular to all vectors $y \in \mathcal{M}$. Then, it can be shown that every vector $x \in \mathcal{H}$ has a unique decomposition $x = x\_{\mathcal{M}} + x\_{\bot}$ with $x\_{\mathcal{M}} \in \mathcal{M}$ and $x\_{\bot} \in M^{\bot}$. Based on this decomposition, one defines the mapping $P\_{\mathcal{M}} : \mathcal{H} \rightarrow \mathcal{H}$ by

$P\_{\mathcal{M}}(x) = x\_{\mathcal{M}} , x \in \mathcal{H}$

Tasks :

1. Is $P\_{\mathcal{M}}$ linear?
2. What is the range $R(P\_{\mathcal{M}})$ and the nullspace $\mathcal{N}(P\_{\mathcal{M}})$ of this mapping?
3. Determine the norm of $P\_{\mathcal{M}}$
4. Verify that
$P\_{\mathcal{M}}(x) = \arg \min\_{y \in \mathcal{M}} \left\| x - y \right\|\_{\mathcal{H}}$
, i.e. $\hat{x} = P\_{\mathcal{M}} x$ is the vector that minimizes $\left\| x - y \right\|\_{\mathcal{H}}$ among all $y \in \mathcal{M}$.

**My thoughts so far :**

1. It is easy to see that $P\_{\mathcal{M}}(\alpha x\_1 + \beta x\_2) = (\alpha x\_1 + \beta x\_2)\_{\mathcal{M}} = \alpha \cdot (x\_1)\_{\mathcal{M}} + \beta \cdot (x\_2)\_{\mathcal{M}}$ via the properties of the linear subspace $\mathcal{M} \subset \mathcal{H}$.
2. The orthogonal projection is a projection for which the range and nullspace are orthogonal subspaces. Hence, we have that the range $R(P\_{\mathcal{M}}) = \mathcal{M}$ as defined above. Consequently, the nullspace must be $\mathcal{N}(P\_{\mathcal{M}}) = M^{\bot}$.
3. The norm can be derived as follows. Consider $\left\| Px \right\|^2 = |\langle Px, Px \rangle| = |\langle x, Px \rangle| \leq \left\| x \right\| \cdot \left\| Px \right\|$. Accordingly, we have that $\left\| P \right\| = \frac{\left\| Px \right\|}{\left\| x \right\|} \leq 1$
4. By inspection, it becomes clear that $y$ is the solution to the problem, since perpendicularity implies that the scalar product in the corresponding space is zero. But how to verify that??

I would appreciate a quick look at my results as well as an intuition as how to verify task 4. Any help is appreciated :-)