Help with this simple example about projection of linear spaces

Question

I'm studying the book Linear Algebra by Hoffman and Kunze and I'm trying to find a concrete example in $\mathbb R^2$ of this fact:

If $R$ and $N$ are sub­spaces of $V$ such that $V=R \oplus N$, there is one and only one projection operator $E$ which has range $R$ and null space $N$. That operator is called the projection on $R$ along $N$.

I'm thinking about the linear space

$$V=\langle(1,1)\rangle$$

being $E:V\to V$ the projection given by

$$E(x,y)=(x,0)$$

Following my calculations we should have $R=\langle(1,0)\rangle$ and $N=\langle (0,1)\rangle$. I don't understand why

$$R\oplus N=\mathbb R^2\neq V$$

Answer 1

Note that neither $R$ nor $N$ are actually subspaces of $V$ in your example and your suggested projection, considered from $V\to\mathbb{R}^2$ has trivial kernel and range of $R$. As @hardmath noted in their comment, this result is much more interesting to think about in the case where $V$ has dimension $>1$.

Answer 2

The map $E$ you defined doesn't map to your $V$, it's rather a map $\Bbb R^2\to\Bbb R^2$, which would justify $R\oplus N=\Bbb R^2$.

On the proof of the claim: by the direct sum property, every vector $v\in V$ can be uniquely decomposed as $v=r+n$ with $r\in R,\ n\in N$. Now define $$P(r+n):=r$$ By the uniqueness of the decomposition, you can verify that it's linear, and obviously one has $P^2(r+n)=r=P(r+n)$, hence $P^2=P$.

On the other hand, if $P^2=P$ and $\ker P=N,\ {\rm range\,}P=R$, then we must have $Pn=0$ for all $n\in N$ and for any $r\in R$ there's an $x\in V$ with $P(x)=r$, then $$P(r)=P(P(x))=P^2(x)=P(x)=r\,,$$ so, $P$ must satisfy $P(r+n)=r$.