How to visualize the orthogonal complement in a vector space of two dimensions? Suppose I have a vector space $V$ of two dimensions, i.e., $V \subseteq \mathbb{R}^2$. Then, I define the column space to be $\mathcal{C}(V) = \{w: w = \sum_{i=1}^{2}c_i v_i\}$ where $c_i \in \mathbb{R}$. 
Now, suppose that my orthogonal complement space is defined to be:
$$
\mathcal{C}^{\perp}(V) = \{w: <w,y> = 0, \ \ \forall \ \mathcal{C}(V)\} = \{w: w^{T}V = 0\}
$$
The way I now visualize the column space is that on a 2-D space I have $v_1$ pointing one direction and $v_2$ pointing another, and that the space between them is the column space. Now, I am not sure how the orthogonal complement should look. Is it a three-dimensional vector poking everywhere orthogonally out of the column space? How should I view it? Thanks!
 A: First, note that vector spaces don't have a corresponding column space. Typically, only matrices have column spaces (defined to be all possible linear combinations of the matrix's column vectors).
To answer your question, it depends on the dimension of the column space.


*

*If $\dim \mathcal C(A) = 0$ so that $\mathcal C(A)$ only contains the zero vector, then $\dim (\mathcal C(A))^\perp = 2$ so that $(\mathcal C(A))^\perp$ is the entire plane $\mathbb R^2$.

*If $\dim \mathcal C(A) = 1$ so that $\mathcal C(A)$ is a line, then $\dim (\mathcal C(A))^\perp = 1$ so that $(\mathcal C(A))^\perp$ is the line perpendicular to the previous line, intersecting at the origin.

*If $\dim \mathcal C(A) = 2$ so that $\mathcal C(A)$ is the entire plane $\mathbb R^2$, then $\dim \mathcal C(A) = 0$ so that $\mathcal C(A)$ only contains the zero vector.

A: We speak about orthogonal complement of a subspace of a vector space. 
Since $V$ is already of dimension $2$, then $V^\perp$ has dimension $0$ in $\mathbb{R}^2$.
More generally, in $\mathbb{R}^n$, for every subspace $W$ of $\mathbb{R}^n$, $\dim (W)+\dim (W^\perp)=n$ by rank nullity theorem with the orthogonal projection on $W$.
