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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.

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1 Answer 1

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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.

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  • $\begingroup$ Could you give more details that if $f_i$'s are linearly independent then image of $T$ is whole space. Would be very grateful! $\endgroup$
    – RFZ
    Dec 3, 2019 at 19:28
  • $\begingroup$ 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 $x_m=\sum_{j=1}^{m-1} c_j w_j$. $\endgroup$
    – P Vanchinathan
    Dec 4, 2019 at 2:30
  • $\begingroup$ Still cannot get you. And what does it mean? $\endgroup$
    – RFZ
    Dec 4, 2019 at 4:37
  • $\begingroup$ What is $w_j$?? $\endgroup$
    – RFZ
    Dec 4, 2019 at 4:38
  • $\begingroup$ @ZFR $w_j$ is a typo. It should be $x_j$ Thanks for pointing it out. $\endgroup$
    – P Vanchinathan
    Dec 5, 2019 at 0:06

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