This is a very basic question about understanding dimension (and codimension).
I was given that the definition for codimension is:
Let $W\subset X$ be a subspace. Then $\operatorname{codim} W=\dim(X\setminus W)$.
But there is a definition on a book saying that:
Given a linear space $X$, a proper linear subspace $M$ is called a linear space of codimension one if for a given $x_{0}\in X\setminus M$, every $x\in X$ can be represented as in the form
$$x=\alpha x_{0}+y$$ where $y$ is a scalar and $y\in M$.
And also, I have seen that many people use this representation (and its uniqueness) to show that the codimension of continuous linear non-zero functional is 1.
But I don't quite understand why this implies $\operatorname{codim}=1$ based on the definition in the very beginning? I think that I need to compute the dimension of the set of $x_{0}$, that is $x_{0}=(x-y)/\alpha$. Then I try to compute the basis for this, but then I get stuck.
Could someone explain this to me or help me understand dimension in a more sensible way?
Thanks a lot!
Best Answer
If you have a subspace $W\in X$ of co-dimension $1$, the definition of co-dimension as $\dim(X)-\dim(W)$ means that you can write $X=W\oplus Z$, with $Z$ of dimension $1$.
Then every vector $x$ in $X$ can be written as $x=z+y$ for some $z\in Z$, $y\in W$. But since $Z$ has dimension $1$, if you pick an arbitrary non-null $x_0\in Z$, every $z\in Z$ can be written as $a\cdot x_0$ for some scalar $a$, and thus $x$ can be written as $ax_0+y$.
Conversely, if every vector in $x\in X$ can be written as $x=ax_0+y$, with $y\in W$ and $x_0\notin W$, then taking any basis of $W$, and adding $x_0$, yields a set of $\dim(W)+1$ linearly independent vectors (they are linearly independent because you can't obtain $x_0$ as a linear combination of the others!); $W$ must then have dimension equal to $1$ less than the space they generate (i.e. $X$).