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Linearly independent

  • Thread starter Dustinsfl
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  • #1
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Let x1, x2,...,xk be linear independent vectors in a vector space V.

If we delete a vector, say xk, from the collection, will we still have a linearly independent collection of vectors? Explain.

By deleting a vector from linearly independent span, the other vectors, I believe, will remain independent; however, I don't know how to prove it.
 

Answers and Replies

  • #2
Dick
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Why don't you start with the definition of linearly independent? That's usually a good strategy.
 
  • #3
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Well vectors are lin. ind. if the the det doesn't 0 and if all coefficients are 0. Since I know they are already ind., deleting one shouldn't change the coefficients but I don't know how to set it up in a proof still.
 
  • #4
Dick
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det of what? I don't think that has much to do with the definition of linearly independent. Does it? I suggest you look it up. State it clearly.
 
  • #5
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Determinant of the vectors in the span doesn't equal 0 then they are linearly ind.
 
  • #6
Dick
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Baloney. Determinant is only defined for a square matrix. That's a special case. There's a much more general definition of linear independence.
 
  • #7
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What does it mean if x1, x2, x3 are linearly independent? It means that the solution to a1x1 + a2x2 + a3x3 = 0 is ai = 0 for all i=1,2,3. Apply this definition to k vectors.

Now, does this still hold if you take out some vector in {x1,...., xk}? Remove some xi from the set and construct the equation I did above. Does it follow that all the ai's are 0?
 
  • #8
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If you remove a vector, the other coefficients should still remain the same = 0.
 
  • #9
HallsofIvy
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Yes, and therefore what?

VeeEight is suggesting a proof by contradiction.
 

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