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Linear Independence and Intersections of Sets

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Let E' and E'' be linearly independent sets of vectors in V. Show that [itex]E' \cap E''[/itex] is linearly independent.

    3. The attempt at a solutionTo show a contradiction, let [itex]E' \cap E''[/itex] be linearly dependent. Also let A be all of the vectors in [itex]E' \cap E''[/itex]. Thus, [itex]A \subseteq E'[/itex] and [itex]A \subseteq E''[/itex]. Because A is linearly dependent, there exists [itex]A_1,...,A_n[/itex] distinct vectors in A such that

    [itex]a_1 A_1 + ... + a_n A_n = \vec0[/itex], where [itex]a_1,...,a_n[/itex] are not all zero. But if such a nontrivial linear combination of vectors in A exists, then E' must be linearly dependent since [itex]A \subseteq E'[/itex]. But this is contrary to our definition that E' is linearly independent. This is similarly contradictory for E''. Thus, it is shown that [itex]E' \cap E''[/itex] cannot be linearly dependent and must rather be linearly independent.

    First of all, I don't know if this proof is correct (although it seems conceivable to me). Also, I didn't know how to prove the problem statement directly, so I had to do it by contradiction. If anyone could give me a hint as to how to do this directly, I would be grateful.
     
  2. jcsd
  3. Jan 14, 2012 #2
    Your proof seems to do the job, though I recommend you consider the case when [itex]E'\cap E''[/itex] is empty separately first.

    If you have some theorems at your disposal then you can shorten up your proof considerably by simply noting that a non-empty subset of a linearly independent set is itself linearly independent.
     
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