orthogonal subspaces in $\mathbb R^2$ Here is the question:
Let W be a subspace of $R^n$ with an orthogonal basis {$w_1$,...,$w_p$}, and let {$v_1$,...,$v_q$} be and orthogonal basis for W$\perp$. Assume that both p >= 1 and q>=1.
(1) For the case when n=2, describe geometrically W and W$\perp$. In particular, what are the possible dimensions of these two subspaces?
(2) For the case when n=3.....
My confusion is because I was under the impression that an orthogonal basis must consist of at least two vectors, because one vector cannot be orthogonal alone. But, if W is a two dimension subspace, then how can W$\perp$ be in $R^2$? If W is 2 dimensions, then the only orthogonal subspace would be a line in $R^3$. It seems to me that if both W and W$\perp$ are an orthogonal basis for a subspace, they must each span 2 vectors and together reside in a higher dimension than $R^2$. I also know that the union set of these two subspaces must span $R^2$ by definition, which tells me that there must be a total of two orthogonal vectors. Can someone give me a hint on what I'm missing here?
Thanks
 A: If W is a space of two dimensions then it is not a subspace of R^2 it is the whole space.
The space perpendicular to W is null.
A: $\mathbb{R}^2=W\oplus W^\perp$ (read about direct sums and orthogonal complements ) so $\dim(\mathbb{R}^2)=\dim(W)+\dim(W^\perp)$. Since $\dim(\mathbb R^2)=2$, consider the two cases for $\dim(W)$ and $\dim(W^\perp)$:


*

*$\dim(W)=0$ and $\dim(W^\perp)=2$

*$\dim(W)=1$ and $\dim(W^\perp)=1$ 

*$\dim(W)=2$ and $\dim(W^\perp)=0$


But you were told $p\geq 1$ and $q\ge1$ so the first and third are ruled out. Hence, $\dim(W)=\dim(W^\perp)=1$. 
Geometrically, these are two lines in $\mathbb R^2$. For example, envision $\mathbb R^2$ as $W\oplus W^\perp$ where $W:=\{(x,0):x\in\mathbb R\}$ and $W^\perp:=\{(0,y):y\in\mathbb R\}$. These are just the two lines in $\mathbb R^2$ formed by the $x$ axis and $y$ axis in standard coordinates.

(2) For the case when n=3.....

I'm not sure what your question is. Is it to find the dimensions of $W$ and $W^\perp$? If so, case it out just as above, but this time knowing the dimensions sum to $3$. Then either 


*

*$\dim(W)=3$ and $\dim(W^\perp)=0$

*$\dim(W)=2$ and $\dim(W^\perp)=1$ (Think: $xy$ plane and $z$ axis.)

*$\dim(W)=1$ and $\dim(W^\perp)=2$ (Think: $x$ axis and $yz$ plane.)

*$\dim(W)=0$ and $\dim(W^\perp)=3$.


Again the $p,q\geq 1$ condition rules out the first and last of these. The middle two cases are (geometrically) a line and a plane in $\mathbb R^3$.

My confusion is because I was under the impression that an orthogonal
  basis must consist of at least two vectors, because one vector cannot
  be orthogonal alone.

What if the space is 1 or 0 dimensional? Then no basis for that space can have two vectors in it.
