Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Restriction of SO(N) to 2 dim subspaces

  1. May 18, 2010 #1
    If [itex]N>2[/itex] and [itex]A\in\textrm{SO}(N)[/itex] are arbitrary, does there exist subspaces [itex]V_1,V_2\subset\mathbb{R}^N[/itex] such that

    [tex]
    V_1+ V_2 = \mathbb{R}^N,\quad\quad \textrm{dim}(V_1)=2,\quad \textrm{dim}(V_2)=N-2
    [/tex]

    and such that the restriction of [itex]A[/itex] to [itex]V_1[/itex] belongs to [itex]\textrm{SO}(2)[/itex], and that the restriction of [itex]A[/itex] to [itex]V_2[/itex] is the identity [itex]1[/itex]?
     
  2. jcsd
  3. May 18, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If V is a nonzero subspace, then its image under SO(N) is all of RN.
     
  4. May 18, 2010 #3
    You must have paid too much attention to the title of my post, which I did not write very carefully.

    In the actual problem I'm interested in finding some subspaces with fixed matrix [itex]A\in\textrm{SO}(N)[/itex].
     
  5. May 18, 2010 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ah, my bad. Have you tried looking at the eigenvalues and eigenvectors of A?
     
  6. May 19, 2010 #5
    I don't know anything else about the eigenvalues except that they have norm 1.

    The situation with eigen equation of a orthogonal matrix is like this:

    [tex]
    Az=\lambda z,\quad A\in\textrm{SO}(N),\quad z\in\mathbb{C}^N,\quad \lambda\in\mathbb{C},\quad |\lambda|=1
    [/tex]

    I know how to prove the claim for N=3. I guess it is often considered trivial... "all rotations have some axis". IMO it's not trivial. At least my own proof for it is not trivial, but complicated... :blushing: It involves such special trickery that it cannot be generalised to N>3.
     
  7. May 19, 2010 #6
    Okey, here's a counter example to my own claim

    [tex]
    A = \frac{1}{\sqrt{2}}\left(\begin{array}{cccc}
    1 & 1 & 0 & 0 \\
    -1 & 1 & 0 & 0 \\
    0 & 0 & 1 & 1 \\
    0 & 0 & -1 & 1 \\
    \end{array}\right) \in \textrm{SO}(4)
    [/tex]

    duh...
     
  8. May 19, 2010 #7
    New problem!

    Let [itex]N=1,2,3,\ldots[/itex] and [itex]A\in\textrm{SO}(2N+1)[/itex] be arbitrary. Does there exist a one dimensional subspace [itex]V_1\subset\mathbb{R}^{2N+1}[/itex] such that the restriction of [itex]A[/itex] to [itex]V_1[/itex] is the identity, and the restriction to [itex]V_1^{\perp}[/itex] is in [itex]\textrm{SO}(2N)[/itex]?
     
  9. May 19, 2010 #8
    hmhm.... I only now started thinking about diagonalizability of an orthogonal matrix.

    If I want to prove that a symmetric matrix is diagonalizable, I would use the facts that a matrix always has at least one eigenvalue, and then the symmetric matrix remains symmetric in orthogonal transformation. Using these facts the proof can be carried out inductively.

    Also an orthogonal matrix remains orthogonal in orthogonal transformations, so the diagonalizability of an orthogonal matrix can be proven precisely in the same way as the diagonalizability of a symmetric matrix?

    The answer to may latest question seems to be positive. It can be proven by using eigenbasis of the orthogonal matrix. So all rotations in odd (2N+1) dimensional spaces are always rotations in even (2N) dimensional subspace, around some one dimensional axis?
     
  10. May 19, 2010 #9

    lavinia

    User Avatar
    Science Advisor

    [tex] \left(\begin{array}{cc}0&-1\\1&0\end{array}\right) [/tex]

    This matrix is orthogonal but not diagonalizable - over the reals.
    Over the complexes it is just the complex number, i
     
    Last edited: May 19, 2010
  11. May 19, 2010 #10
    It is well known that real matrices can have complex eigenvalues, so diagonalizable usually does not mean diagonalizable in real vector space only.

    It is not a complex number [itex]i[/itex] unless it is specifically and explicitly defined to be a complex number through some identification [itex]\mathbb{C}\subset\mathbb{R}^{2\times 2}[/itex].

    The standard answer is that that matrix is diagonalizable and its eigenvalues are [itex]i[/itex] and [itex]-i[/itex], with eigenvectors

    [tex]
    \left(\begin{array}{c}
    1+i \\ 1-i \\
    \end{array}\right),\quad\quad
    \left(\begin{array}{c}
    i+1 \\ i-1 \\
    \end{array}\right)
    [/tex]

    That means that your matrix is similar to a diagonal matrix

    [tex]
    \left(\begin{array}{cc}
    i & 0 \\
    0 & -i \\
    \end{array}\right)
    [/tex]
     
  12. May 19, 2010 #11

    lavinia

    User Avatar
    Science Advisor

    I was responding to the thought that an orthogonal matrix can be shown to be diagonalizabe in the same way that a symmetric matrix can. The matrix I provided is a counter example.
    You can not diagonalize this matrix with orthogonal transformations.

    I was just pointing out that multiplication by i, which is the same as this matrix, has no one dimensional real subspaces. Perhaps it would have been clearer if is called it rotation by 90 degrees. the choice of orientation of the plane and the basis that I chose was obvious from the example.
     
  13. May 19, 2010 #12
    I see. My comment that orthogonal matrices remain orthogonal in orthogonal transformations is not enough for the proof, because unitary transformations are going to be needed.

    The claim that orthogonality would be preserved under unitary transformations seems to be incorrect, if I checked it correctly now.

    I think that the fact that a unitary matrix remains unitary under unitary transformations is what is needed. If we use it, we can prove that unitary matrices are diagonalizable in the same way as we can prove that hermitian matrices are diagonalizable too.

    Then, orthogonal matrices are unitary, so the diagonalizability of orthogonal matrices follows.
     
  14. May 19, 2010 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I believe this to be true -- you can (I'm pretty sure) show that eigenvalues that aren't 1 come in pairs -- either (-1,-1) or a complex-conjugate pair. I'm not 100% sure, though -- I can't seem to recall why I believe that to be true.
     
  15. May 19, 2010 #14
    [itex]A\in\textrm{SO}(2N+1)[/itex] has odd number of eigenvalues, and they are on the circle [itex]|z|=1[/itex] so that those that are not [itex]\pm 1[/itex] come in pairs [itex](\lambda,\lambda^*)[/itex]. So the amount of eigenvalues [itex]\pm 1[/itex] has to be odd. The product of each pair [itex](\lambda,\lambda^*)[/itex] always produces [itex]1[/itex], so the number of eigenvalues [itex]\lambda=-1[/itex] has to be even too. The number of eigenvalues [itex]\lambda=1[/itex] has to be odd then, and cannot be zero.
     
    Last edited: May 19, 2010
  16. May 19, 2010 #15

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Oh right, the norm 1 thing; that's what I couldn't remember today. I feel better now. :smile: I was getting worried I gave some bad advice to someone recently!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Restriction of SO(N) to 2 dim subspaces
  1. Subspaces R^n (Replies: 8)

Loading...