Proving a set of functionals is independent. A functional is a linear transformation from an $n$-dimensional vector space $V$ to its scalar field $F$.
I need to prove that a set of functionals $\{ f_1,....,f_n \}$ is independent (in the linear algebra sense) if and only if the intersection of the kernels of $f_i$ for each $1\leq i \leq n$ is $\{ 0 \}$.
Would appreciate some hints.
 A: Assume that $f_1,\ldots,f_n$ are linearly independent, and let $x_1,\ldots, x_n$ a basis of $V$ dual to the $f_i$'s, i.e. $f_i(x_j)=\delta_{ij}$.  
$x\in \bigcap_{j=1}^n\mathrm{ker}\,f_j$, $x\ne 0$, then $x=c_1x_1+\cdots+c_nx_n$, where not all $c_i$'s are equal to zero, and hence
$$
0=f_i(x)=\sum_{i}^n c_jf_i(x_j)=c_i,
$$
which is a contradiction.
Assume now that $\bigcap_{j=1}^n\mathrm{ker}\,f_j=\{0\}$ and $c_1f_1+\cdots+c_nf_n=0$, for $c_i$'s not all zero. Without loss of generality assume that
$$
f_n=d_1f_1+\cdots+d_{n-1}f_{n-1}, \quad d_1,\ldots,d_{n-1}\in \mathbb F.
$$
In such case 
$$
\bigcap_{j=1}^{n-1}\mathrm{ker}\,f_j\subset\mathrm{ker}\,f_n,
$$ 
and hence
$$
\bigcap_{j=1}^{n-1}\mathrm{ker}\,f_j=\bigcap_{j=1}^{n}\mathrm{ker}\,f_j.
$$
But $\mathrm{ker}\,f_j$ is a $(n-1)$-dimensional subspace (if $f_j\ne 0$) and thus $\bigcap_{j=1}^{n-1}\mathrm{ker}\,f_j$ is of dimension at least $1$. Thus $\bigcap_{j=1}^{n}\mathrm{ker}\,f_j\ne\{0\}$.
A: Your $n$ functionals together define a linear map $f:V\to F^n$, whose kernel is the intersection of the kernels of the $f_i$. Having a linear relation  $a_1f_1+\cdots+a_nf_n=0$ between $f_1,\ldots,f_n$ means that the linear form $a:F^n\to F$ with matrix $(a_1~a_2~\ldots~a_n)$ satisfies $a\circ f=0$, in other words the kernel of $a$ contains the image of$~f$. Linear independence of $f_1,\ldots,f_n$ then means that this only happens when the linear form $a$ is itself $0$, and this is the case if and only if the image of $f$ is all of $F^n$, or equivalently when $f$ has rank $n$.
So your question translates into: show that $f$ has rank $n$ if and only if $\dim\ker(f)=0$. Since it was given that $\dim V=n$, this is an immediate consequence of the rank-nullity theorem.
A: Note that the rank of functionals is 1. Also in the finite dimensional set up, the functionals are nothing but inner product with some vector. (as V of dim n ~ $\mathbb{K}^n$).  
So let there be $v \neq 0$ in the intersection of kernels. Consider the orthogonal subspace of the linear subspace generated by this vector. It is (n-1) dimensional call it W. 
$ f_i(x) = \langle v_i , x \rangle $ for some $ v_i \neq 0 $ note $v_i \not\in$ Ker $f_i$ .
Now, functionals are independent if $v_i$s are independent. But we see that we have to find n linearly independent vectors from a (n-1)-dimensional space W, which is impossible as every n vectors in such a space will be linearly dependent.
And the other direction follows from the fact that $v_i$s form a basis of $V$ when $f_i$ are linearly independent.    
