Let $V$ be a linear space s.t. $\dim\, V=n$ and $f_1,f_2,\dots,f_m$ be linear functionals ($m
Let $V$ be a linear space s.t. $\dim\, V=n$ and $f_1,f_2,\dots,f_m$ be
  linear functionals ($m<n$). Show that $\exists x\neq 0$ s.t.
  $f_i(x)=0$ $\forall i$.
We can suppose that these $f_i$ are linearly indepent.
Let $\{e_i\}_{i=1}^n$ be a basis to $V$ and $\{\delta_i\}_{i=1}^n$ be its dual basis.
We can also set $f_i=\sum_{k=1}^n\alpha_{ik}\delta_i$. Then, $f_i(e_j)=\alpha_{ij}$
Let $x\in V$, then $x=\sum_{j=1}^n \beta_j e_j$.
Therefore, we have that $f_i(x)=\sum_{j=1}^n\beta_j\alpha_{ij}$.
And I'm stuck here.

$F = (f_1,\ldots,f_m)$ defined by $F(x) = (f_1(x),\ldots,f_m(x))$ is a linear map from $V$ to $F^m$.
$\dim(\ker(F)) + \dim(\operatorname{im}(F)) =  \dim(V)$. 
We know that the right hand side equals $n$. The image of $F$ has dimension at most $m < n$. So the dimension of $\ker(F)$ cannot be $0$, so has some vector $x \neq 0$ in it. This $x$ works.
