Question :
Show that if $f_1 , f_2, \ldots ,f_m $ are linear forms on a n-dimmensional vector space $V$ over $K$, then $\dim \ \cap_{i=1}^m \ker(f_i) \geq n-m$ with equality iff $f_1 , f_2, \ldots , f_m$ are linear independent.
I have tried :
Define an Linear transformation $T : V \rightarrow K^m $ such that $T(v) = \begin{bmatrix} f_1(v)\\ f_2(v)\\ \vdots \\ f_m(v) \end{bmatrix}$
$\therefore$ $\ker (T) = \cap_{i=1}^m \ker(f_i)$, using rank nulity theorem ,we get
$ n = \ker (T) + rank(T) \leq \ker (T) + m $ $\Rightarrow \ \cap_{i=1}^m Ker(f_i) \geq n-m $
Please tell me how to show that the equality hold iff $f_1 , f_2 , \cdots ,f_m$ are linear independent.