A fundamental fact about Linear Algebra

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SUMMARY

The discussion centers on proving that a vector space $V$ over an infinite field $F$ cannot be expressed as a set-theoretic union of a finite number of proper subspaces. The proof employs a contradiction method, assuming that such a union exists and deriving a contradiction using the Pigeonhole Principle (PHP). The conclusion is that the assumption leads to a contradiction, confirming the original statement as true.

PREREQUISITES
  • Understanding of vector spaces and their properties
  • Familiarity with infinite fields in linear algebra
  • Knowledge of the Pigeonhole Principle (PHP)
  • Basic proof techniques, particularly proof by contradiction
NEXT STEPS
  • Study the properties of vector spaces over infinite fields
  • Learn more about the Pigeonhole Principle and its applications in proofs
  • Explore advanced topics in linear algebra, such as dimension theory
  • Review proof techniques, focusing on contradiction and direct proofs
USEFUL FOR

This discussion is beneficial for mathematics students, particularly those studying linear algebra, as well as educators and researchers looking to deepen their understanding of vector space properties and proof strategies.

caffeinemachine
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MHB
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Hello MHB,
This is probably my first challenge problem which falls in the 'University Math' category.

$V$ is a vector space over an infinite field $F$, prove that $V$ cannot be written as a set theoretic union of a finite number of proper subspaces.
 
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Nobody participated :(

Here's my solution:

Assume contradictory to the problem. Let $n$ be the minimum integer such that $V$ can be written as $V=V_1\cup\cdots\cup V_n$ where each $V_i$ is a proper subspace of $V$. Thus, \begin{equation*}\forall i,\exists x_i\in V \text{ such that } x_i\in V_j\iff j=i\tag{1}\end{equation*}Now consider $S=\{f_1x_1+\cdots+f_nx_n:f_i\in F\}$. Clearly this set is infinite, thus, by PHP, there is a $k$ such that $a,b\in V_k$ for distinct $a$ and $b$ in $S$. This contradicts $(1)$. Hence we achieve the required contradiction and the proof is complete.
 
caffeinemachine said:
Nobody participated :(

I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.
 
MarkFL said:
I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.
My bad then. Next time I'll wait for a week.
 

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