linear functionals linearly independent

Question

Let $V$ be a vector space with $\dim V=n$. Let $\varphi_1,...,\varphi_n $ be linear functionals that are not $0$. Prove that $\varphi_1,...,\varphi_n $ are linearly independent if and only if $\cap_{i=1}^n \ker \varphi_i = \{0\}$. $\\$ I succeeded to prove that if they are linearly independent then the intersection is zero but I have no idea how to prove the other direction. Any suggustion? thanks!

Answer 1

Note that we can define a map from $V$ to $\Bbb F^n$ by $$ \Phi(x) = \pmatrix{\varphi_1(x)\\ \vdots \\ \varphi_n(x)} $$ If the intersection of their kernels is $0$, then the map define by $\Phi$ is invertible. That is, $\Phi(x) = Ax$ for an invertible matrix $A$ (with respect to some basis).

For any $c_1,\dots,c_n$, note that $$ c_1 \varphi_1(x) + \cdots + c_n\varphi_n(x) = c^T A x $$ where $c$ is the column vector of $c_i$. What we want to show then is that $$ c^T A = 0 \implies c = 0 $$ which follows from the invertibility of $A$.