Proof: Max number of Linearly Independent Vectors

  • #1
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Homework Statement


Prove that a set of linearly independent vectors in Rn can have maximum n elements.

So how would you prove that the maximum number of independent vectors in Rn is n?


I can understand why in my head but not sure how to give a mathematical proof. I understand it in terms of the number of independent vectors being equal to the rank of the matrix they create and obviously a matrix of dimension n can only have max n pivots. But I don't think that's really sufficient for a proof.
 
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Answers and Replies

  • #2
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But I don't think that's really sufficient for a proof.
If you write it down properly, that should be fine.
The main point: a n x m matrix (for your m vectors) cannot have rank larger than n.
 
  • #3
Ray Vickson
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Homework Statement


Prove that a set of linearly independent vectors in Rn can have maximum n elements.

So how would you prove that the maximum number of independent vectors in Rn is n?


I can understand why in my head but not sure how to give a mathematical proof. I understand it in terms of the number of independent vectors being equal to the rank of the matrix they create and obviously a matrix of dimension n can only have max n pivots. But I don't think that's really sufficient for a proof.

Yes, it is (with a bit of tweaking). If vectors ##v_1, v_2, \ldots, v_n, v_{n+1} \in R^n## are linearly independent, we should not be able to find ##(c_1, c_2, \ldots, c_{n+1}) \neq(0,0, \ldots, 0)## giving ##c_1 v_2 + c_2 v_2 + \cdots + c_n v_n + c_{n+1} v_{n+1} = \vec{0}##. Assuming, instead, that you CAN find such ##c_i##, you should be able to get a contradiction.
 
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