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
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=\frac13(1,1,1)+\frac13(1,-1,0)+\frac13(1,0,-1),$$you must have$$A.(1,0,0)=\frac13(1,-1,0)+\frac13(1,0,-1)=\left(\frac23,-\frac13,-\frac13\right).$$So, the entries of the first column of the matrix of $\operatorname{proj}_V$ with respect to the standard basis will be $\frac23$, $-\frac13$ and $-\frac13$. Can you take it from here?