$\pi$ self-adjoint
$\iff \forall x, y \in V, \langle \pi(x)\mid y\rangle=\langle x\mid \pi(y)\rangle$
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle \pi(x_U+x_W)\mid y_U+y_W\rangle=\langle x_U+x_W\mid \pi(y_U+y_W)\rangle$
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U+y_W\rangle=\langle x_U+x_W\mid y_W\rangle$
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$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle+\langle x_W\mid y_W\rangle=\langle x_U\mid y_W\rangle+\langle x_W\mid y_W\rangle$
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$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle=\langle x_U\mid y_W\rangle$
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$\iff \forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0$
Let's restrict our attention to subspaces $V$ of $\mathbb{R}^3$ rather than $\mathbb{R}^n$. Once this case is understood, you can try to generalize it. It is important to think slowly from the definitions. Geometric intuition will come afterwards (and be correct). I will not recall the definition of orthogonal projection onto a subspace for you, you can look that up in your notes/textbook.
You appear to be confusing several concepts. Let me try to clarify them for you.
Fix a subspace $V \subseteq \mathbb{R}^3$ (this could be the origin, a line through the origin, a plane containing the origin, or the entire space $\mathbb{R}^3$). Let $T_V\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the linear transformation defined by orthogonal projection onto the subspace $V$.
Any linear transformation has a kernel and an image. They are defined for $T_V$ as follows:
$$\text{image}(T_V) = \left\{ y \in \mathbb{R}^3 \colon \exists x \in \mathbb{R}^3 \text{ such that } T_V(x) = y \right\} $$
$$\text{kernel}(T_V) = \left\{ x \in \mathbb{R}^3 \colon T_V(x) = 0 \right\}$$
(you may note that both the image and the kernel of $T_V$ are subspaces of $\mathbb{R}^3$).
From the first definition, we can explain that $$\text{image}(T_V) = V.$$ The proof uses two key facts: the definition of the image of a linear transformation, and the definition of the map $T_V$.
Proof that $\text{image}(T_V) = V$: In order to do this, we show that $\text{image}(T_V) \subseteq V$ and $V \subseteq \text{image}(T_V)$:
For any vector $x \in \mathbb{R}^3$, the orthogonal projection of $x$ onto $V$ is an element of $V$. Thus $\text{image}(T_V) \subseteq V$.
On the other hand, if $x$ is an element of $V$, then $T_V(x) = x$ (the orthogonal projection of a vector in $V$ onto $V$ is itself), so $V\subseteq \text{image}(T_V)$. This completes the proof. $\square$
Thus:
- If $V$ is a line in $\mathbb{R}^3$, then $\text{image}(T_V)$ is the same line.
- If $V$ is a plane in $\mathbb{R}^3$, then $\text{image}(T_V)$ is the same plane.
- If $V$ is the entire space $\mathbb{R}^3$, then $\text{image}(T_V)$ is the entire space $\mathbb{R}^3$.
Now we would like to describe the second space, $\text{kernel}(T_V)$. In order to do this, it is useful to recall that the orthogonal complement of a subspace $V$ is a new subspace defined in the following way:
$$V^{\perp} = \left\{ y\in \mathbb{R}^3 : \forall x\in V, \langle x,y\rangle = 0 \right\}.$$
In plain English, $V^{\perp}$ is the set of all vectors that are orthogonal to every vector in $V$.
You should think about why the following statements are true (note that tehy only make sense if $V$ is a subspace of $\mathbb{R}^3$):
- If $V$ is a line, then $V^{\perp}$ is a plane.
- If $V$ is a plane, then $V^{\perp}$ is a line.
- If $V$ is the origin, then $V^{\perp}$ is the entire space $\mathbb{R}^3$.
- If $V$ is the entire space $\mathbb{R}^3$, then $V^{\perp}$ is the origin.
You should also try to draw pictures of some examples.
The following statement contains the intuition you are after. I will leave the proof of this to you.
$$V^{\perp} = \text{kernel}(T_V).$$
Best Answer
Take $x \in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=\xi$ for some $\xi \in \operatorname{Ker}P$. Thus $x = Px + \xi$. This shows that $V=\operatorname{Im}P + \operatorname{Ker}P$. Now take $y \in \operatorname{Im}P \cap \operatorname{Ker}P$. Since $y \in \operatorname{Im}P$ we have $y=Pz$ for some $z \in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y \in \operatorname{Ker}P$, hence $0=Py=P^2z=Pz=y$. This shows that $\operatorname{Im}P \cap \operatorname{Ker}P=\{0\}$ and so we have $V=\operatorname{Im}P \oplus \operatorname{Ker}P$.