Suppose $V$ is an infinite dimensional vector space, $f_i$ ($i$ is from $1$ to $n$) are real-valued linear functions on $V$, I cannot understand why the intersection of kernels of $f_i$ must contain a nonzero element. If this question can be answered, then it is easy to settle my problem. Thanks.
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1The co-dimension of a non-trivial linear functional is $1$. See this. – David Mitra Mar 18 '14 at 08:21
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Of the kernel, that is. – David Mitra Mar 18 '14 at 08:59
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I got it. Thank you for your kind help. – S.C. Liu Mar 18 '14 at 23:44
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Suppose that the intersection of the kernels contains only the zero vector. Consider $n+1$ vectors $v_1,\ldots, v_{n+1}$. We can pose the homogenous linear system $$ f_j(v_1)c_1+\cdots+f_j (v_{n+1})c_{n+1}=0, \ j=1,\ldots,n $$ of $n$ equations in the $n+1$ variables $c_1,\ldots c_{n+1}$. This system has a nontrivial solution and so $$ f_j(c_1v_1+\cdots+c_{n+1}v_{n+1})=0, \ j=1,\ldots,n, $$ which implies that $c_1v_1+\cdots+c_{n+1}v_{n+1}$ is in the intersection of the kernels. So $c_1v_1+\cdots+c_{n+1}v_{n+1}=0$, showing that $v_1,\ldots,v_{n+1}$ are linearly dependent. Thus, $\dim V\leq n$.
Martin Argerami
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