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Rotation matrix proof

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Let A be 2x2 and det(A)=1 and entries in R. Suppose A does not have any real eigenvalues. Then prove there exists a B st B is 2x2, det(B)=1 and BAB^-1=[cos(x),sin(x),-sin(x),cos(x)]
    for some x.




    3. The attempt at a solution
    I'm not sure how to start this proof. Any help would be appreciated.
     
  2. jcsd
  3. Nov 24, 2009 #2

    Dick

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    If A is real and doesn't have real eigenvalues then it has two complex conjugate eigenvalues and their product is 1. That means they can be written in the form cos(x)+/-i*sin(x). Can you say why all these things are true? Isn't that starting to sound like a rotation matrix? Can you pick B to change the standard basis into a linear combination of those eigenvectors so that you get that form for BAB^(-1)?
     
  4. Nov 24, 2009 #3
    I understand the part about the eigenvalues being complex conjugates (I can use the quadratic formula to find the eigenvalues of an arbitrary 2x2 matrix, and then I get that their product is 1), but I don't understand how to get B.
     
  5. Nov 24, 2009 #4

    Dick

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    Did you use the fact that det(A)=1?
     
  6. Nov 24, 2009 #5
    yes, that made the computation much easier.
     
  7. Nov 24, 2009 #6
    I guess I'm not getting it. The eigenvector depends on the eigenvalue. If A=[a,b,c,d] then the eigenvalue is [(a+d)+-sqrt((a+d)^2-4)]/2. If I plug this in and try to find an eigenvector, it gets messy fast. I end up with an equation where an eigenvector is [0,1] and d-a-sqrt((a+d)^2-4)-2cb/[a-d-sqrt((a+d)^2-4)] must equal 0 for the matrix to be singular. What am I supposed to derive from this?
     
  8. Nov 24, 2009 #7

    Dick

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    You are doing this WAY TOO explicitly. You want to show the matrix exists, not explicitly calculate it. You can think of B as just representing a choice of basis. You know (I hope) that you can write the two eigenvalues as cos(x)+isin(x) and cos(x)-isin(x) for some x. Can you explain why? So let v1 and v2 be the corresponding two eigenvectors. That means if you pick the B corresponding to that choice of basis, then BAB^(-1) is diagonal with the two eigenvalues along the diagonal. Can you think of a way to make linear combinations of v1 and v2 to create another basis where A takes the given form?
     
    Last edited: Nov 24, 2009
  9. Nov 24, 2009 #8
    Thanks for helping me; sorry I'm a little slow on the uptake. I'll just repeat the parts that I understand. Ok, so the fact that det(A)=1 means that the eigenvalues are the two conjugates you mentioned, I get that. Then we can associate those with 2 eigenvectors that form a basis. How do we know that BAB^-1 is a diagonal matrix? By "corresponding", do you mean using the eigenvectors as the columns of B?
     
  10. Nov 24, 2009 #9

    Dick

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    Yes, that's it. In this case I would use the eigenvectors as the columns of B^(-1), Then B^(-1) maps (1,0) and (0,1) into v1 and v2. A multiplies each one by it's eigenvalue and B transforms the eigenvectors back to (1,0) and (0,1) times the eigenvalues. Hence the product is diagonal. Now do some more messing around with the basis to get the form you want.
     
    Last edited: Nov 24, 2009
  11. Nov 24, 2009 #10
    Ok, but A is the original matrix, not the one with eigenvalues on the diagonal. Is that what you mean?
     
  12. Nov 24, 2009 #11

    Dick

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    Look. Here's an example. The matrix A=[[2,1],[1,2]] has eigenvalues 1 and 3. The eigenvectors are simple. Construct a matrix B such that BAB^(-1)=[[1,0],[0,3]]. Just do it for practice, ok? No, A is not diagonal. Now suppose I told only that the matrix A has eigenvalues 1 and 3 without telling what the matrix is. You can still say if I set the matrix B^(-1) to have columns v1 and v2 where v1 has the eigenvalue 1 and v2 has the eigenvalue 3 then BAB^(-1) is [[1,0],[0,3]]. Do you see why? Once you've done that with the eigenvalues cos(x)+/-i*sin(x) you still have to rearrange the basis a bit to get the indicated form for BAB^(-1).
     
  13. Nov 25, 2009 #12
    Ok, thanks for the simple example. I think I get the concept now. Any tips on how to manipulate the B's? I need to find a P st PBAB^-1P^-1 is the rotation matrix, right? Does the restriction that det(PB)=1 matter?
     
    Last edited: Nov 25, 2009
  14. Nov 25, 2009 #13

    Dick

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    det(B)=1 doesn't make it any more difficult. If C=BAB^(-1) then C=(B*k)*A*(B*k)^(-1) for any constant k. I've been trying to explain that you shouldn't worry about the form of B. What you should be trying to do is find a basis where A takes the form you want. Then describe the B. It just does the basis change.
     
  15. Nov 25, 2009 #14
    ok, so we want a basis st A acts as a rotation. I'm having trouble conceptualizing what basis would do that.
     
  16. Nov 25, 2009 #15

    Dick

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    You know a basis that diagonalizes A, right? Try this. Diagonalize [cos(x),sin(x),-sin(x),cos(x)]. You'll get the same diagonal matrix. Now work backwards.
     
  17. Nov 25, 2009 #16

    Dick

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    Try this. Take the matrix [cos(x),sin(x),-sin(x),cos(x)] and diagonalize it. You'll get the same eigenvalues as A. Now figure out the relation between the basis where A looks like a rotation matrix and the basis of eigenvectors.
     
  18. Nov 25, 2009 #17
    ok, I see that I can turn A into the rotation matrix form using B=LP, where A=PDP^-1 and the rotation matrix=LDL^-1. Is that correct? however, I still would need to show that B is real. Without knowing the eigenvectors of A, but knowing the eigenvectors of the rotation matrix, how do I do that?
     
  19. Nov 25, 2009 #18

    Dick

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    You ask a lot of questions without showing a lot of work, no offence, but what does B=LP mean? And why do you think think B has to be real? I'm going to ask you once more. What are the eigenvalues and eigenvectors of [[cos(x),sin(x)],[-sin(x),cos(x)]]? If you can actually work through this that might give you some insight into how to handle the case were the real matrix A has det(A)=1 and has no real eigenvalues.
     
    Last edited: Nov 26, 2009
  20. Nov 26, 2009 #19
    Let R be the rotation matrix you mentioned for some x. then R=BAB^-1, where A can be diagonalized as PDP^-1 and R can be diagonalized as LDL^-1. Then D=P^-1AP and R=LP^-1APL^-1.
    So B=LP^-1, since, given A, we can find an R with the same eigenvalues as A. the eigenvalues of R are cosx+-isinx and the respective eigenvectors are [+-i,1]. I apologize if I seem like I'm not trying to do any work on this; I did work out the example you gave before, I just did not mention it. Anyway, I hope I'm being clear on what I've got so far. The last thing is to show B is real, since that was part of the problem.
     
    Last edited: Nov 26, 2009
  21. Nov 26, 2009 #20

    Dick

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    Ok. The trouble is that I'm not getting enough feedback to tell where you are in the proof? Can you summarize your progress so far? What I really wanted out of that is that the eigenvectors are v1=[-i,1] and v2=[i,1]. In that basis A is diagonal. Notice that the original basis where A looks like a rotation can now be written (0,1)=(v1+v2)/2 and (1,0)=(v2-v1)/(2i). Can you apply that to your problem?
     
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