Prove that the following set is Linearly Dependent 
Let $n$ be a positive integer and let $V$ be a vector space of dimension $n$. Let $S =\{f_1, \cdots, f_n\}$ be a subset of $V^∗$ and assume that there exists a non-zero vector $v$ in $V$ satisfying $f_i(v) = 0$ for all $1 \le i \le  n$. Show that $S$ is linearly dependent.

Here's how I've started...
Assume set $S$ is Linearly Independent, since $\dim(V)=n$, it will be a basis for $V^*$. Now I find an element which cannot be spanned by $S$.
How to proceed after this?
The question has been answered here, but we yet haven't been taught transposes of linear transformations and therefore is a bit tough for me to understand.
 A: Let's start from where you stopped: "Assume set S is Linearly Independent, since dim(V)=n, it will be a basis for V*."
Since the $f_i$'s vanish on $v$, any linear combination of them, i.e. any element of $V^*$, will also vanish on $v$.
But this is impossible since $v\ne0:$ take a basis of $V$, the "coordinate functions in that basis" are linear forms on $V$, and one of them does not vanish on $v.$
A: Assume towards contradiction, $S$ is independent. Since $|S|$ $=n$ $=\mathrm{dim}(V)$ $=\mathrm{dim}(V^*)$, we have $S=\{f_1,…,f_n\}$ is basis of $V^*$. So $\mathrm{span}(S)=V^*$. By hypothesis, $\exists 0\neq v\in V$ such that $f_i(v)=0$, $\forall i\in J_n$. So $\{v\}$ is independent. It’s easy to check $\exists v_2,v_3,…,v_n\in V$ such that $\{v,v_2,…,v_n\}$ is basis of $V$. It’s a standard fact $\exists !f\in V^*$$=L(V,F)$ such that $f(v)=1_F$ and $f(v_i)=0_F$, $\forall 2\leq i\leq n$. Since $f\in V^*$$=\mathrm{span}(\{f_1,…,f_n\})$, we have $\exists !a_1,…,a_n\in F$ such that $f=\sum_{i=1}^na_i\cdot f_i$. So $f(v)$ $=(\sum_{i=1}^na_i\cdot f_i)(v)$ $= \sum_{i=1}^na_i\cdot f_i(v)$ $=0_F$ $=1_F$. Thus we reach contradiction.
This problem is very similar (in terms of idea) to this post.
