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I Question about decomposition of matrices in SL(2,R)

  1. Jul 5, 2016 #1
    Hello,

    we are given a 2×2 matrix [itex]S[/itex] such that [itex]det(S)=1[/itex].
    I would like to find a 2x2 invertible matrix [itex]A[/itex] such that: [itex]A S A^{-1} = R[/itex], where [itex]R[/itex] is an orthogonal matrix.

    Note that the problem can be alternatively reformulated as: Is it possible to decompose a matrix SSL(2,ℝ) in the following way: [tex]S=A^{-1}R A [/tex]where R is orthogonal and A is invertible?

    Is this a well-known problem? To be honest, I don't have many ideas on how to tackle this problem, so even a suggestion that could get me on the right track would be very welcome.
     
  2. jcsd
  3. Jul 5, 2016 #2

    micromass

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    Every orthogonal matrix can be diagonalized. Not every matrix in ##SL(2,\mathbb{R})## can be diagonalized. So what you ask is impossible.
     
  4. Jul 5, 2016 #3

    fresh_42

    Staff: Mentor

    You may count the numbers of degrees of freedom you have in either matrix.
     
  5. Jul 5, 2016 #4
    Why did you mention the diagonalizability of a matrix?
    I am not asking if S is similar to a diagonal matrix. I am rather asking if S is similar to an orthogonal matrix.
     
  6. Jul 5, 2016 #5

    fresh_42

    Staff: Mentor

    ##R## orthogonal ⇒ ##R## diagonalizable ⇒ (if ##R## similar to ##S##) ##S## diagonalizable
    ##S = \begin{pmatrix} 1 && 1 \\ 0 && 1 \end{pmatrix} \in SL(2,ℝ)## not diagonalizable, contradiction
    ⇒ ##S## not similar to ##R##
     
  7. Jul 6, 2016 #6
    I thought orthogonal matrices were not diagonalizable over ℝ in general.


    Btw, I actually found a counterexample of your (micromass' and fresh's) above statements: consider the matrix [itex]
    S = \begin{pmatrix}
    1 & 2\\
    -1 & -1
    \end{pmatrix}
    [/itex].
    We can easily see that [itex]det(S)=1[/itex], and we can still decompose that matrix into: [tex]
    S = \underbrace{\begin{pmatrix}
    -1 & -1\\
    0 & 1
    \end{pmatrix} }_{A^{-1}}\;

    \underbrace{\begin{pmatrix}
    0 & -1\\
    1 & 0
    \end{pmatrix}}_R\;

    \underbrace{\begin{pmatrix}
    -1 & -1\\
    0 & 1
    \end{pmatrix}}_A
    [/tex]

    where A is invertible and R is orthogonal, as required. This proves that there exist matrices in SL(2,ℝ) that are similar to orthogonal matrices. Note also that S is not diagonalizable over ℝ.
     
    Last edited: Jul 6, 2016
  8. Jul 6, 2016 #7

    micromass

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    They are over ##\mathbb{C}##.

    ##S## is diagnalizable over ##\mathbb{C}##. And of course there exists matrices in ##SL(2,\mathbb{R})## similar to orthogonal matrices. The identity matrix would be an example of this. The point is that not all matrices in ##SL(2,\mathbb{R})## are similar to orthogonal matrices.
     
  9. Jul 6, 2016 #8

    micromass

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    Furthermore, what you ask is possible exactly for those matrices in ##SL(2,\mathbb{R})## whose eigenvalues have ##|\lambda|=1## (and could be complex).
     
  10. Jul 7, 2016 #9
    Thanks micromass,

    I think I understand your remarks, but I still have a doubt.

    Given a matrix [itex]\mathrm{S} \in SL(2,\mathbb{R})[/itex], and assuming there exists a 2×2 real matrix matrix A such that [itex]A^{-1}SA = R[/itex], we can deduce that S must be diagonalizable (over ℂ), since every rotation matrix R is. That means that we can write: [tex](AC)^{-1}\,S\,(AC) = \Lambda[/tex] where we used the diagonalization [itex]R=C\,\Lambda\,C^{-1}[/itex] (note that both C and Λ are in general complex).


    Now, starting from the sole knowledge of S we could perform an eigendecomposition of S and we would obtain [itex]S=Q^{-1}\Lambda Q[/itex], where [itex]Q=AC[/itex]. But then, how do we extract the real matrix A from the complex matrix Q?

    For instance, in the example I gave above where [itex]
    S = \begin{pmatrix}
    1 & 2\\
    -1 & -1
    \end{pmatrix}
    [/itex] the eigendecomposition of S would give [itex]S=Q \Lambda Q^{-1}[/itex] where: [tex]Q=
    \frac{\sqrt{6}}{6}\begin{pmatrix}
    2 & 2\\
    -1+i & -1-i
    \end{pmatrix}
    [/tex]
    [tex]\Lambda=
    \begin{pmatrix}
    i & 0\\
    0 & -i
    \end{pmatrix}
    [/tex]

    How do we extract [itex]A=
    \begin{pmatrix}
    -1 & -1\\
    0 & 1
    \end{pmatrix}
    [/itex] from the complex matrix Q ?
     
    Last edited: Jul 7, 2016
  11. Jul 12, 2016 #10
    Ok, I think this question has been almost answered.

    Summarizing, given a matrix [itex]S\in SL(2,\mathbb{R})[/itex] we ask if there exist one invertible matrix [itex]A\in GL(2,\mathbb{R})[/itex] and a rotation matrix [itex]R\in SO(2,\mathbb{R})[/itex] such that [itex]A^{-1}SA=R[/itex].

    Since [itex]R[/itex] is diagonalizable over ℂ (i.e. [itex]R=C\Lambda C^{-1}[/itex]), we have that:

    [tex]S=(AC)\Lambda (C^{-1}A^{-1}) = Q\Lambda Q^{-1} \qquad\qquad\qquad \mathrm{(1)}[/tex]
    From the above equation we can deduce that [itex]S[/itex] must be diagonalizable and must have eigenvalues having modulo 1. These are necessary conditions. They would be also sufficient conditions if we add the requirement that the eigenvectors of [itex]S[/itex] must have the same real part.

    Since [itex]Q=AC[/itex] we can find a matrix [itex]A=QC^{-1}[/itex], where the columns of [itex]C[/itex] contain the eigenvectors of [itex]R[/itex], for instance [itex]C=\begin{pmatrix}
    1 & 1\\
    i & -i
    \end{pmatrix}
    [/itex].
    It can be verified that when the eigenvectors of [itex]S[/itex] have the same real part, then the matrix [itex]A[/itex] is real.

    However, the matrix [itex]A[/itex] is not unique. For example, we can easily see that by multiplying [itex]A[/itex] with an invertible matrix [itex]M[/itex] that commutes with [itex]S[/itex] (i.e. [itex]MS=SM[/itex]) Equation (1) would still hold.
     
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