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.

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.
