Question on Inner Product Space

Question

I know that an inner product space can be split into 2 vector spaces $W$ and $W^ \perp$. So if $x$ is not in $W ^\perp$. However, when I think about it, if I have a line that passes through origin in $\mathbb R^2$ space, say spanned by $(1,1)$ and call it $L$. Then if I take a point that is not on $L$, say $(1,2)$, then it must be in $L^ \perp$. However, I can't see how $\langle (1,1),(1,2)\rangle $ will give me $0$. What am I misunderstanding here? The proof of $V = W \oplus W^\perp$ is convincing but somehow I can't reconcile that fact with this example.

Answer 1

"Orthogonal complement" and "complement" are not the same. In your example, the orthogonal complement of $ (1,1) $ is the span of $(-1,1) $. And then $$(1,2)=\frac32\, (1,1)+\frac12\, (-1,1). $$ The vector $(1,2) $ is in the complement of $ L $, but not in its orthogonal complement.

Answer 2

If $V = W \oplus W^\perp$, that does not mean that every element of $V$ is either in $W$ or in $W^\perp$. It means every element of $V$ can be split into its $W$ component and its $W^\perp$ component – that is, it can be written as $x + y$ with $x \in W$ and $y \in W^\perp$, and moreover there's only one $x$ and only one $y$ satisfying these conditions, which justifies me calling them "components".

In your example, $\langle (1,1) \rangle^\perp = \langle (1,-1)\rangle$, so what I'm saying here is I can write $(1,2)$ as $(1.5, 1.5) + (-0.5, 0.5)$, and those two vectors fall in your $W$ and $W^\perp$ respectively.

Since you asked in a different comment how we know $\langle (1,1) \rangle^\perp = \langle (1,-1) \rangle$, well, we know that $(1,-1)$ is orthogonal to $(1,1)$ by direct computation, and then we know that any scalar multiple of $(1,-1)$ will be orthogonal to any scalar multiple of $(1,1)$, so we know the orthogonal complement of $\langle (1,1) \rangle$ at least includes $\langle (1,-1) \rangle$. Since we know from the theorem you quoted that 2D space is equal to the direct sum of this space and its orthogonal complement, and the dimension of the direct sum is the sum of the dimensions of the components, we know the dimensions of both of those must be $1$, so there's nothing else in the complement.