$f:V\to F$ a linear functional, $N=\ker f$, $v_0\in V-N\implies\exists!(c,n)\in F\times N$ s.t. $v=cv_0+n\ \forall v\in V$

Question

We have a linear functional $f:V\to F$ on a vector space $V$ over a field $F$.

$N=\ker f$, $f\not=0,\ N\not=V.$

And $v_0\in V-N$ is fixed.

I want to show that there uniquely exist $c\in F$ and $n\in N$ such that $v=cv_0+n$ for all $v\in V$.

My work so far:

1. $v\in N\subset V\implies$ Let $c=0\in F,\ n=v\in N$ and we have unique $c$ and $n$ such that $v=cv_0+n=0\cdot v_0+v.$

2. $v\in V-N\implies$ Let $c=\frac{f(v)}{f(v_0)}\in F$ (since $v_0\in V-N\implies f(v_0)\not=0$), then we have $$f(v)=f(cv_0)$$ since $f$ is linear, and $$v=cv_0+n$$for all $n\in N.$

The problem is that at the step 2, $c$ is determined uniquely but $n$ is not. And I sort of can guess that $$f(v)=f(cv_0)\implies v=cv_0+n$$ for all $n\in N$ is true, but couldn't construct a rigorous proof. (maybe it is false?)

Answer 1

Actually in the second step you are missing one little point that would give you the solution: We know that if $v=cv_0 +n$ then $f(v) = f(cv_0+n) = cf(v_0)$ which gives us the uniqueness of $c$ as $c= \frac{f(v)}{f(v_0)}$ (which is well-defined in all cases becasue $v_0 \notin N$).

Now we know that $f(v) = f(cv_0+n)$ for any $n \in N$, but that does NOT mean $v= cv_0 + n$ for any $n \in N$ because we don't know if $f$ is injective. So we still have to find a unique $n$. But this is easy: If we were to have $v= cv_0 +n$ this would imply $v-cv_0 = n$. Therefore: Just set $n:= v-cv_0$ with the $c$ found above. Then we have $f(n) = f(v) - f(cv_0)=0$ so indeed $n \in N$ and this is the unique $n$ such that $v=cv_0 +n$.