What is the significance of estimation space and error space in linear models? Consider a linear model $E(y)=X\beta,\,\operatorname{Var}(y)=\sigma^2I_n$ where $y$ is an $n\times 1$ response vector, $\beta$ is a $p\times 1$ vector of parameters and $X$ is an $n\times p$ design matrix with $\operatorname{rank}(X)=r\le p\,(<n)$.
In the book Plane Answers to Complex Questions by Ronald Christensen, there is this concept of estimation space and error space in relation to a linear model. Estimation space is defined as the column space of $X$ and error space is defined as the orthogonal complement of the estimation space, i.e. the null space of $X^T$. I am trying to understand the significance of this definition. Does the estimation space represent the vector space of all estimable linear functions of $\beta$? What does the error space represent? Can I say that it is the vector space of all functions $C^Ty$ such that $E(C^Ty)=0$? I can see that dimension of estimation space is $r$ and dimension of error space is $n-r$.

I'd like to add to frank's answer. Estimation space is the column space of $X$, denoted $\mathcal{C}(X)$. The space of estimable linear functions of $\beta$ is, in a sense, the row space of $X$, i.e., $\mathcal{C}(X^T)$. They are not the same. By the definition of estimability, we have that $$c'\beta\ \text{is estimable}\iff c\in\mathcal{C}(X^T).$$
In addition, there are only $r=\mathrm{rank}(X)$ linearly independent estimable linear functions.
Just like orthogonally decomposing a vector in $\mathbb{R}^n$ onto the estimation space $\mathcal{C}(X)$ and the error space $\mathcal{N}(X^T)$, we can decompose a vector in $\mathbb{R}^p$ onto $\mathcal{C}(X^T)$ and $\mathcal{N}(X)$. Suppose $c'\beta$ is estimable. Then $c\in\mathcal{C}(X^T)$. For any $u=u_1+u_2$ with $u_1\in\mathcal{C}(X^T)$ and $u_2\in\mathcal{N}(X)$, we can see that
$$c'u=c'(u_1+u_2)=c'u_1$$
because every vector in $\mathcal{C}(X^T)$ is orthogonal to every vector in $\mathcal{N}(X)$. Geometrically speaking, to find a solution to the linear system $X^TX\beta=X^Ty$, we can first project $y$ onto $\mathcal{C}(X)$ to get $\hat{y}$, which is equal to $Xu$ for some $u\in\mathbb{R}^p$, and then find a specific solution $u_1\in\mathcal{C}(X^T)$ such that $Xu_1=\hat{y}$. Such $u_1$ is unique, because $\mathcal{C}(X^T)$ and $\mathcal{C}(X)$ has the same dimension and hence are $1$-to-$1$ corresponded. Hence, a solution for $\beta$ is $\beta^0=u_1+u_2$, for any $u_2\in\mathcal{N}(X)$. It follows that $c'\beta^0=c'u_1$ is unique.
