Dual space with $W=\ker T$ Let $f \in V^*$ with $V$ a vector space and $W=ker f $. If $v_0 \in V$ is a vector such that $f(v_0)\ne 0$ then for every $v \in V$ there exist unique $w \in W$ and scalar $c$  such that $v=cv_0+w$.
How can I prove this I don't understand it, please if someone can help me. Thanks for you time and help.
 A: Since there exists $v_0 \in V$ such that $f(v_0) \neq 0$, we know $f$ is not trivial. Since $V$ is a vector space over a field $F$, and $f:V \rightarrow F$ is linear, we must then have $f$ surjective, since $F$ has no nontrivial proper subspaces (this is a property of fields). Then we have by the first isomorphism theorem: $V/ \ker(f) \cong F$, so then $V/\ker(f)$ is a one dimensional vector space, i.e. for every $v \in V$ there is a $c \in F$ such that $v+ \ker(f)=cv_0+\ker(f)$. But then $v+h=cv_0+k$ for some $h,k \in \ker(f)$, so letting $w=k-h$, $w \in \ker(f)$ and $v=cv_0+w$, as desired. 
If you don't want to use the first isomorphism theorem, use rank-nullity:
$\dim(im(f))+\dim(\ker(f))= \dim(V)$, so $\dim(\ker(f))= \dim(V)- 1$, i.e. $\dim(V/\ker(f))=1$ and the same result follows.
A: There are great answers already, but here's another argument. 
It's nice to work backwards here. If $v=cv_0+w$, what does that get you? Set
$$c = \frac{f(v)}{f(v_0)}\quad\text{and}\quad w = v-\frac{f(v)}{f(v_0)}v_0.$$
In this case, $v=cv_0+w$, and $w\in\ker{f}$. Suppose that $v=c'v_0+w'$ is another representation. Then, $f(c'v_0+w')=f(cv_0+w)$ which implies 
$$c'f(v_0) = cf(v_0),$$
and therefore we must have $c=c'$. It follows that $w=w'$ as well, giving uniqueness.
A: Let $n$ be the dimension of $V$.
It is an application of the classical direct sum decomposition $V=W \oplus W^{\perp}$ where $W$ is $n-1$ dimensional (an hyperplane, the kernel of linear form $f$) and $W^{\perp}$ is the one-dimensional subspace directed by $v_0$, thus $W^{\perp}=\mathbb{R}v_0$. Now use the fact that a direct sum means unique decomposition.
