Basis Proof

  • Thread starter pyroknife
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  • #1
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I attached the problem and its solution.

I was looking at this solution and got a little confused. Why did they say that "Assume that S = {v1, v2, · · · , vn} is a basis for V and c is a nonzero scalar. Let S1 = {cv1, cv2, · · · , cvn}. Since S is a basis for V , V has dimension n. Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S is independent."

That seems redundant. The problem statement already stated that S is a basis, which means that the vectors in S are linearly independent.
 

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  • #2
Dick
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I attached the problem and its solution.

I was looking at this solution and got a little confused. Why did they say that "Assume that S = {v1, v2, · · · , vn} is a basis for V and c is a nonzero scalar. Let S1 = {cv1, cv2, · · · , cvn}. Since S is a basis for V , V has dimension n. Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S is independent."

That seems redundant. The problem statement already stated that S is a basis, which means that the vectors in S are linearly independent.

It's clearly a minor typo. They meant to say, "Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S1 is independent."
 
  • #3
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It's clearly a minor typo. They meant to say, "Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S1 is independent."

Oh okay. So by saying that, wouldn't that have proved S1 is a basis? The next part (writing out all the linear combinations) just proved that it's linearly independent again. Do you need to do that?
 
  • #4
Dick
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Oh okay. So by saying that, wouldn't that have proved S1 is a basis? The next part (writing out all the linear combinations) just proved that it's linearly independent again. Do you need to do that?

Sure you do. That's the proof part!
 

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