# How to establish isomorphism between kernel of linear functional and scalar field?

So, I have to show that the kernel of a linear functional is a subspace of codimension one. Someone suggested to me that the following:

If $$f:X\rightarrow \mathbb{K}$$ and $$f\not\equiv 0$$, then $$\ker(f)$$ is a subspace of codimension one and a subspace of codimension one is kernel of non-zero linear functional.

To prove this, we can do the following:

$$f$$ induces an isomorphism between $$X\backslash \ker(f)$$ and $$\mathbb{K}$$. Conversely, if $$Z$$ is a hyperplane, let $$g:X\rightarrow X\backslash Z$$ be the natural map and let $$T:X\backslash Z\rightarrow \mathbb{K}$$ be an isomorphism. Then, $$f=T\circ g$$ is a linear functional on $$X$$ and $$\ker(f)=Z$$.

But my problem is: how do I establish those isomorphisms? Can somebody help me out here? Thanks in advance.

To check that $$f=T\circ g$$ is a linear functional you simply apply the definitions. Finally $$x\in \ker f$$ iff $$T(g(x))=0$$, but $$T$$ is an linear functional so this occurs iff $$g(x)=0$$, but $$g(x)=0$$ iff $$x\in Z$$ by the definition of the natural quotient map.
If you were wondering how to construct $$T$$: By axiom of choice there exists a basis $$\{e_i\}$$ for a $$Z$$. We can append to that an element $$e$$ so that $$\{e\}\cup \{e_i\}$$ is a basis for $$Z$$. Define $$T':X\to \mathbb K$$ by extending $$\alpha e\mapsto\alpha$$ by Hahn-Banach. Define $$T:X\backslash Z\to \mathbb K$$ in the natural way from $$T'$$. I leave it to you to show that $$T$$ is an isomorphism. This is actually the standard way to show the converse.
Note we are assuming that $$\dim X>1$$ here, else the result is trivial.