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T/F: Orthogonal matrix has eigenvalues +1, -1

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    If a 3 x 3 matrix A is diagonalizable with eigenvalues -1, and +1, then it is an orthogonal matrix.

    2. Relevant equations


    3. The attempt at a solution
    I feel like this question is false, since the true statement is that if a matrix A is orthogonal, then it has a determinant of +1 or -1, which has nothing to do with diagonalozation. However, I don't see how to prove this rigorously. Would the best way just be to search for a counter-example?
     
  2. jcsd
  3. Oct 27, 2016 #2

    micromass

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    What do you know about the eigenvectors of an orthogonal matrix?
     
  4. Oct 27, 2016 #3
    I can't seem to recall anything about the eigenvectors of an orthogonal matrix... I looked it up and couldn't find anything either.
     
  5. Oct 27, 2016 #4

    fresh_42

    Staff: Mentor

    How is an orthogonal matrix defined? And how do eigenvalues behave in that equation?
     
  6. Oct 28, 2016 #5
    An orthogonal matrix is a square matrix such that ##A^{T}A= I##. I don't see how this can be used to analyze its eigenvalues. I can only see that detA = 1 or -1... But those aren't eigenvalues.
     
  7. Oct 28, 2016 #6

    fresh_42

    Staff: Mentor

    Firstly, we have to show (or know), that eigenvalues of ##A^ {T}## are the same as those of ##A##.
    This is done by the equation ##\det (M)=\det(M^T)## and therewith ##\det(\lambda I -A^T)=\det((\lambda I-A^T)^T)= \det(\lambda I^T - (A^T)^T)=\det(\lambda I - A)##

    Secondly, for an eigenvector ##x## of ##A## with an eigenvalue ##\lambda## what do we get from ##Ax = \lambda x\,##?

    Now what does it mean for ##A## to be diagonalizable, and what must be the values of this diagonal?
     
  8. Oct 28, 2016 #7
    If A is diagonlizable, then it can be represented in the form A = PDP-1, where P is a matrix of linearly independent eigenvectors of A, and D is a diagonal matrix with the eigenvalues of A on its diagonal. But I'm not seeing how this definition helps me.

    Also, I'm not sure what you mean by what do we get from the ##Ax = \lambda x\##. The only thing I see to do with that might be to multiply both sides by the transpose of A and see what happens...
     
  9. Oct 28, 2016 #8

    fresh_42

    Staff: Mentor

    I meant the consideration of a orthogonal matrix ##A##. Being orthogonal leads to ##x=I\, x=A^TA\, x = A^T(A(x))=\lambda^2 x## for an eigenvector ##x## of the (orthogonal) matrix ##A## to the eigenvalue ##\lambda##. Thus ##\lambda^2 = 1## and over the real numbers this means ##\lambda \in \{-1,+1\}##.

    So orthogonal matrices must have eigenvalues ##\pm 1##.

    Now we have the opposite situation: The eigenvalues are given as ##\pm 1##, which is a necessary condition. Checked.
    Then we have that ##A## is diagonalizable, i.e. ##A=PDP^{-1}## for some matrix ##P##.
    The diagonal entries of ##D## also have to be ##\pm 1## by the given condition, which means ##D=D^T=D^{-1}## and ##A=A^{-1}##.
    Thus to show ##A^TA=I## is equivalent to show ##A^T=A## or ##A^T=A^{-1}##.

    Here is where I'm stuck. To solve for the corresponding linear equations is rather unpleasant. (I'll answer, if I find the trick.)
     
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