# Linear Form on a n-dimensional vector space $V$

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.

Rank of $T$ is the same as the dimension of the image (range) of $T$. The image is contained in an $m$ dimensional vector space. To check that it is exactly $m$-dimensional is the same condition that $f_i$'s are linearly independent.

• Could you give more details that if $f_i$'s are linearly independent then image of $T$ is whole space. Would be very grateful!
– RFZ
Dec 3, 2019 at 19:28
• Equality of two functions means, their values are same at every point in the domain. In case the the linear forms $f_j$'s are linearly dependent we get an equality of them as function as below: $f_m =\sum_{j=1}^{m-1}c_jf_j$. Now equate their values at every $v\in V$. We see that in the image every vector $w=(x_1,x_2,\ldots, x_m)$ satisfies $x_m=\sum_{j=1}^{m-1} c_j w_j$. Dec 4, 2019 at 2:30
• Still cannot get you. And what does it mean?
– RFZ
Dec 4, 2019 at 4:37
• What is $w_j$??
– RFZ
Dec 4, 2019 at 4:38
• @ZFR $w_j$ is a typo. It should be $x_j$ Thanks for pointing it out. Dec 5, 2019 at 0:06