Linear Independence and Intersections of Sets

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SUMMARY

The discussion centers on proving that the intersection of two linearly independent sets of vectors, E' and E'', is also linearly independent. The proof by contradiction demonstrates that if the intersection E' ∩ E'' were linearly dependent, it would lead to a contradiction regarding the linear independence of E' and E''. Additionally, participants suggest considering the case of an empty intersection separately and utilizing the theorem that states a non-empty subset of a linearly independent set remains linearly independent to simplify the proof.

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Homework Statement


Let E' and E'' be linearly independent sets of vectors in V. Show that [itex]E' \cap E''[/itex] is linearly independent.

The Attempt at a Solution

To 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.
 
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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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