# How to understand codimension?

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!

• $\operatorname{codim}W=\dim (X/W)$ (the quotient space), not $\dim X\setminus W$, which isn't even defined, since $X\setminus W$ isn't a subspace. – Bernard Feb 28 '17 at 12:08

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. $$X$$).
• @J.Y yes, that's right! $\{x_0\}$ is a basis for $Z$. – Anonymous Feb 28 '17 at 12:05