The union of kernels is a proper subspace

Asked Apr 12 '13 at 23:34

Active Apr 13 '13 at 11:25

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I was asked by someone about the confusion when reading the following proof:

Let $V$ be a vector space over a field $F$ of characteristic $0$, and let $f_1,f_2,\ldots, f_n$ be linear mappings from $V$ to $F$ none of them being the zero mapping. Then there exists $a \in V$ so that $f_i(a) \neq 0$ for $i=1,\ldots, n$.

And the proof goes something like this: Take the union of the kernels of the $f_i$. It claims that (which I found confusing) this union is a proper subset of $V$, so there must exist such an $a$ as stated in the question. Can someone give me some explanation of what is going on here? Any help is appreciated.

edited Apr 13 '13 at 11:25

asked Apr 12 '13 at 23:34

user63279

4

Your question is answered in this math.SE thread. – Zev Chonoles Apr 12 '13 at 23:36
thank you very much! @ZevChonoles – user63279 Apr 12 '13 at 23:45
@ZevChonoles I am reading the proof by Steve D on MO: http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces, and I am not sure how $F$ being infinite is used to imply that some $V_j$ contains $y$, can you help me with this? Thanks a lot! – user63279 Apr 13 '13 at 00:24
@YACP Thank you very much! – user63279 Apr 13 '13 at 01:01

The union of kernels is a proper subspace

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