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Show this is a basis.

  1. Oct 27, 2012 #1
    Show that if S = {v1, v2, . . . , vn} is a basis for Rn
    and A is an n × n invertible matrix, then
    S' = {Av1,Av2, . . .,Avn} is also a basis.

    I need to show that:
    1) Av1, Av2,...Avn are linearly independent
    2) span(S)=Rn

    I'm having some problems with this.
    I see that S'=AS (duh)

    This is what I'mthinking of doing:
    c1*A*v1+...cn*A*vn=0 where c1,....cn are constants and must be zero
    =A*c1*v1+...A*cn*vn=0 can I multiply by A^-1 to both sides? If so then:
    =c1v1+...+cn*vn=0=span(S)
    thus S' is a basis?
     
  2. jcsd
  3. Oct 27, 2012 #2

    haruspex

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    That's the sort of thing, but your wording is all wrong. What do you mean by "c1,....cn are constants and must be zero"?
    Start with a stated supposition, like 'elements of S' are not linearly independent'. Therefore there exist, etc.
     
  4. Oct 27, 2012 #3
    well if they form a basis then the linear combination:
    c1v1+...cnvn=0 can only have one solution for the constants which is 0. Or do I have it all wrong?
     
  5. Oct 27, 2012 #4

    haruspex

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    That's correct, but it's not what you wrote before. It may be what you had in mind, but the reader cannot tell that.
    Also, you actually wrote c1*A*v1+...cn*A*vn=0, so clarifying what you wrote before with the explanation you've now given, I get:
    if the elements of S' form a basis then the linear combination c1*A*v1+...cn*A*vn=0 can only have one solution for the constants which is 0​

    True, but not a useful place to start. We're trying to prove that they do form a basis. A logical chain starting with the supposition that they do might prove they don't, but it won't prove they do.
    Try reductio ad absurdum: if they are not linearly independent then there exist...
     
  6. Oct 27, 2012 #5
    So wouldn't this:

    c1*A*v1+...cn*A*vn=0 where c1,....cn are constants and must be zero
    =A*c1*v1+...A*cn*vn=0 can I multiply by A^-1 to both sides? If so then:
    =c1v1+...+cn*vn=0=span(S)

    show that the vectors are linearly independent equal to span(S)?
     
  7. Oct 27, 2012 #6

    haruspex

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    I've already explained that is not a meaningful statement. I can only check your logic if you write it out properly.
     
  8. Oct 27, 2012 #7
    How about:

    c1*A*v1+...cn*A*vn=0
    Rearranging: A*c1*v1+...A*cn*vn=0 can I multiply by A^-1 to both sides? If so then:
    Multiplying A^-1 to both sides gives: c1v1+...+cn*vn=0=span(S) since S is defined as a basis, then there are constants c1...cn that must be 0 for ^ the above to be true because v1...vn are linearly independent.

    This shows that S' is linearly independent because c1...cn for c1*A*v1+...cn*A*vn=0 is equal to 0. Also, span(S')=R^n. S' is a basis.
     
  9. Oct 27, 2012 #8

    haruspex

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    You still don't get what I'm on about. Starting with "c1*A*v1+...cn*A*vn=0" means nothing. Try wrapping some words around it:
    "If the elements of S' are not linearly independent then there exist..."
    Later, yes, you can multiply by A-1 and obtain c1v1+...+cn*vn=0. But what is this "=span(S)"? That is certainly not true.
     
  10. Oct 27, 2012 #9
    Oops I forgot to put that, but here:

    If S' is a basis, then S' must be linearly independent, thus if S' is linearly independent then the following needs to be satisfied:
    c1*A*v1+...cn*A*vn=0 where c1=...=cn must be 0.
    Rearranging: A*c1*v1+...A*cn*vn=0 can I multiply by A^-1 to both sides? If so then:
    Multiplying A^-1 to both sides gives: c1v1+...+cn*vn=0
    Since S is a basis, the vectors in S are linearly independent. Span(S)=c1v1+...+cn*vn, where c1...cn are unique.

    This shows that S' is linearly independent because c1...cn for c1*A*v1+...cn*A*vn=0 is equal to 0. Also, span(S')=R^n. S' is a basis.
     
  11. Oct 27, 2012 #10

    haruspex

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    No. If you want to prove some fact X, it is not useful to start with "if X". Start with "if not X". I keep giving you the kick-off point: ""If the elements of S' are not linearly independent then there exist..."
    Carry on from there.
     
  12. Oct 27, 2012 #11
    Why is that even necessary? It's saying the exact same thing. In every proof in my book that was asked to prove something, they prove the thing that was asked. They don't prove that the opposite is not true.
    I don't understand how that is any more "correct" than what I did.
     
  13. Oct 27, 2012 #12

    haruspex

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    In general, you have some given facts, X, and want to prove a fact Y.
    You can start with X and manipulate them to arrive at Y, or, using the "reductio ad absurdum" approach, you start by assuming Y is false, then show this is inconsistent with X being true.
    What you have tried to do is neither of these. You have started by assuming Y is true and seeing what you can deduce from there. You will never prove Y that way.
     
  14. Oct 27, 2012 #13
    That looks fine to me.
     
  15. Oct 27, 2012 #14
    Okay how about this:

    Suppose c1Av1+...+cnAvn=0. This is of the form Av=0 where v=c1v1+...+cnvn.
    Multiplying A^-1 to both sides give v=0. Since v is a set of linearly independent vectors, c1=...=cn=0. Thus Av is linearly independent.
     
  16. Oct 28, 2012 #15

    haruspex

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    Almost there! To make it clear, you really should start as I indicated:
    If the elements of S' are not linearly independent then there exist c1,..,cn, not all zero, such that c1Av1+...+cnAvn=0. Then, using your steps above, you show that c1,..,cn, are all zero, so the supposition that the elements of S' are not linearly independent must have been false.
    Now, can you finish the problem by showing S' spans the space?
     
  17. Oct 28, 2012 #16
    Hmmm, I completely forgot about that I had to prove that spans the space as well.

    I can use the same idea except the RHS is some vector b in Rn

    c1Av1+...+cnAvn=b
    Multiplying both sides by A^-1 gives:
    c1v1+...+cnvn=A^-1*b (on a side note I've been meaning to ask, when you multiply A^-1 to both sides, why does A^-1 end up in front of the vector b and not behind it?)
    since v1....vn form a basis, this is solvable and thus it spans Rn.
     
  18. Oct 28, 2012 #17
    Another question: since it's already proven to be linearly independent, doesn't that mean it's gonna automatically span Rn? I can't think of a situation where it wouldn't when it's linearly independent.
     
  19. Oct 28, 2012 #18

    haruspex

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    Sorry, but you have the argument backwards again. To prove they span the space, start by taking an arbitrary vector of the space then obtain an expression for the vector in the form of c1Av1+...+cnAvn. I.e. you will need to show how to compute the ci.
    You need to break yourself of the habit of launching into equations without thinking about the structure of the proof. Here's a guide:
    - Decide whether you are trying to prove that something exists/can be done or that something does not exist/cannot be done.
    - If you are trying to prove that something exists, you will normally be looking for a way to construct it. (In the spanning proof needed here, a representation of a given vector using the proposed basis.) Non-constructive proofs also arise, and can be exceedingly elegant, but they're in the minority.
    - If trying to prove something does not exist, it is usually easier to use RAA (reductio ad absurdum): start by supposing it does exist then obtain a contradiction. This was appropriate for the linear independence part: suppose there is a linear relationship and show that contradicts the givens.
     
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