Linear algebra proof

  • #1
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Homework Statement


Prove the following: Let V be a vector space and assume there is an integer n such that if (v1, . . . , vk) is a linearly independent sequence from V then k ≤ n. Prove is (v1, . . . , vk) is a maximal linearly independent sequence from V then (v1, . . . , vk) spans V and is therefore a basis.

Homework Equations




The Attempt at a Solution


If v_1,...,v_k$spans V then all vectors in V are generated by some linear combination of v_1,...,v_k. It's clearly seen that we can generate any vector in the sequence by setting the constant of the desired vector to 1 and the others to 0. Hence the sequence is maximal linearly independent, adding another vector will provoke a dependency. The dependency didn't exist before the new vector was added. This implies that the added vector can be written as a linear combination of the other vectors. If we do this for every remaining vector in V, then all vectors can be written as a linear combination of the given sequence and therefore it spans V. The sequence is linearly independent and it spans V, so it's a basis.



is that correct?
 

Answers and Replies

  • #2
You are asked to prove it spans V. Starting with "if v1..vk spans V" is not helpful.
 
  • #3
You are asked to prove it spans V. Starting with "if v1..vk spans V" is not helpful.
So if I delete the first sentence, it would look better? because clearly that's what I want to prove.
 
  • #4
So if I delete the first sentence, it would look better? because clearly that's what I want to prove.
You'll need to delete all that followed from that, i.e. the first two and a half lines. So it now starts
adding another vector will provoke a dependency.
That seems to work.
 

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