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.

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.
