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Linear Algebra orthogonal matrix

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that A is a real n by n matrix which is orthogonal, symmetric, and positive definite. Prove that A is the identity matrix.


    2. Relevant equations
    Orthogonality means [itex]A^t=A^{-1}[/itex], symmetry means [itex]A^t=A[/itex], and positive definiteness means [itex]x^tAx>0[/itex] whenever x is a nonzero vector.


    3. The attempt at a solution
    Messing around with inner products, trying to show that the matrix [itex]A-I[/itex] is the zero matrix. Help is appreciated.
     
  2. jcsd
  3. Aug 27, 2011 #2

    I like Serena

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    Welcome to PF, slamminsammya! :smile:

    Are you aware of the spectral theorem for symmetric matrices?

    What can you say about the eigenvalues of the matrix A?
     
  4. Aug 27, 2011 #3

    micromass

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    Let's prove this is steps. Can you first show that 1 is the only eigenvalue of A?

    Edit: ILS was first :cry:
     
  5. Aug 27, 2011 #4
    Yes, I have been able to show that 1 is the only eigenvalue, but I just cant seem to get from there to showing that A therefore fixes each vector. The proof that 1 is the only eigenvalue goes like this:

    Suppose Ax=cx for a scalar c. Then A(Ax)=c^2x=x, so that the only possible eigenvalues are 1 or -1. But -1 cannot be an eigenvalue, since if that were so then <x, Ax>=-|x|^2 < 0, contradicting positive definiteness.

    From here, wikipedia tells me that from the spectral theorem for symmetric matrices, A must be similar to the identity matrix (since the eigenvalue 1 has multiplicity n). Does this mean that A must be the identity?
     
  6. Aug 27, 2011 #5
    Yes it does... Thanks
     
  7. Aug 27, 2011 #6
    Out of curiosity, it seems like there should be a way of proving this without invoking the spectral theorem for symmetric matrices. Does anyone know a proof?
     
  8. Aug 27, 2011 #7
    I solved this using the Cayley-Hamilton theorem; every square matrix A satisfies it's characteristic polynomial.

    You said At = A, because it's symmetric
    and you've said At = A-1 for orthogonality.

    So combining these results you obtain A = A-1
    and therefore A2 = I, where I is the identity matrix.

    Your minimal polynomial is therefore m(x) = x2 - 1
    The roots of this are your eigenvalues, which are 1 and -1, but we reject -1 as it contradicts positive definiteness. So our eigenvalue is 1, which according to our minimal polynomial has multiplicity 1.
    So this means the jordan normal form of A is the Identity matrix, as the jordan form is n 1's along the diagonal.

    However this only shows that A is similar to I, however, if we are working with the same basis then it surely means A = I...
     
  9. Aug 27, 2011 #8

    I like Serena

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    Nice! :smile:

    For the record, at this moment I would not know how to do it without an advanced theorem.



    It's still not trivial that similarity to the identity matrix implies identity...
     
  10. Aug 27, 2011 #9
    And to think last week I didn't understand it, now it's my first approach to most of these types of questions. :wink:

    Right so we need to use the fact that it's similar to the identity to show that it's equal.

    This is all I can come up with;

    A is similar to I, so A = P-1IP where P is the transition matrix.
    So then, A2 = P-1I2P = P-1IP = A

    But I = A2 = P-1IP = A

    Thus, A = I.


    Seems correct anyway. :smile:
     
    Last edited: Aug 27, 2011
  11. Aug 27, 2011 #10

    I like Serena

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    Yep! Looks correct! :smile:

    Note:
    A is similar to I, so A = P-1I P where P is the transition matrix.
    So A = P-1I P = P-1P = I.
     
  12. Aug 27, 2011 #11

    I completely agree with your method here, but I don't think that x^2-1 is the characteristic polynomial of A. The characteristic polynomial is the determinant of A-cI, and for any n by n matrix the characteristic polynomial must therefore be of degree n. Forgive me if I am misunderstanding, but I don't quite see how Cayley Hamilton applies in the way you were treating it.

    Maybe this is exactly what you were saying, but wouldnt the conclusion be that the characteristic polynomial must be (x-1)^n, and so the only way for A to satisfy this polynomial is if A=I?
     
  13. Aug 27, 2011 #12

    micromass

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    She said that [itex]x^2-1[/itex] is the minimal polynomial (or correctly: that the minimal polynomial divides [itex]x^2-1[/itex]). This is correct.
     
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