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Eigenvectors with at least one positive component

  1. Oct 17, 2012 #1
    I am wondering if there is a systematic way to fix the phase of complex eigenvectors. For example [tex]e^{i \theta}(1,\omega,\omega^2)[/tex] where [itex]e^{i \theta}[/itex] is an arbitrary phase and [itex]\omega[/itex] and [itex]\omega^2[/itex] are the cube roots of unity, is an eigenvector of the cyclic matrix [tex]\left(\begin{matrix}0& 1&0\\0&0&1\\1&0&0\end{matrix}\right).[/tex] But I feel that [itex](1,\omega,\omega^2)[/itex] and equivalently [itex](\omega,\omega^2,1)[/itex] and [itex](\omega^2,1,\omega)[/itex] are somehow special because one of the components is real and positive. Is there some special name for such a choice of phase? Any reference will be greatly appreciated.
     
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  3. Oct 18, 2012 #2

    Erland

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    Re: Eigenvectors with atleast one positive component

    I am not sure what you really want, but any eigenvector has at least one nonzero component, call it a. Dividing the vector with a gives an eigenvector such that (at least) one of its components is 1, which is real and positive.
     
  4. Oct 18, 2012 #3

    AlephZero

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    The phase of an eigenvector is arbitrary, whether it is real or complex.

    If you are finding the solution to some problem as a linear combination of eigenvectors ##\sum_i a_i \phi_i##, you could say that each "component' ##a_k \phi_k## has a definite phase (and magnitude) because the numerical value of ##a_k## depends on the arbitrary choice for the phase of the corresponding ##\phi_k##.

    In some practical situations you might choose to fix the amplitude and phase of an eigenvector in an arbitrary way (e.g. by making the biggest component of the vector = +1) but doing that doesn't have much mathematical significance.
     
  5. Oct 19, 2012 #4
    Thanks guys for the reply. I am aware of the things you pointed out. As Erland said I should have been more specific on what I want. I am studying the representations of discrete groups for application in physics. Eigenvectors of group elements are significant in this context (for example the matrix that I wrote in the earlier post represents a rotation by 2pi/3 about the axis (1,1,1) and the axis (1,1,1) is an eigenvector). The problem of phase arises when I starts looking at complex eigenvectors for example [itex](1, \omega,\omega^2) [/itex]. All I am saying is that a phase convention with one of the components real and positive is "appealing" to me (especially in the physics context) and so I want to know if such a choice is somehow special for a mathematian.
     
  6. Oct 19, 2012 #5
    Ohh, I now notice something. Since my representation matrices are unitary, their eigenvalues are always in the form [itex]e^{i \theta}[/itex]. For the matrix that I wrote in the first post, the eigenvalue (corresponding to the eigenvector [itex](1,\omega, \omega^2)[/itex]) is [itex]e^{i 2\pi/3}[/itex]. So basically the "appealing" phase convention has all its components having phases just a multiple of the eigenvalue.

    So let me make a hypothesis. Please let me know it is true.

    Let [itex]e^{i \theta}[/itex] be an eigenvalue for a special unitary matrix, [itex]U[/itex]. We will always be able to choose a phase convention for the corresponding eigenvector so that all its components have a phase [itex]e^{i n \theta}[/itex] where n is an integer.
     
  7. Oct 19, 2012 #6
    Ok, I was wrong, I tried with random unitary matrices and it didn' t work. So let me modify my hypothesis.

    Let [itex] U[/itex] be a discrete unitary matrix, with [itex] U^n = I[/itex], where n is an integer. Its eigenvalues will be of the form [itex] e^{i m\frac {2\pi} {n}}[/itex] where [itex] m[/itex] is an integer. There exists a basis and a phase convention in which all the components of the eigenvectors are of the form [itex] | a_x | e^{i k_x\frac {2\pi}{n}}[/itex] where [itex] k_x[/itex] are integers.

    Please let me know if the above statement is true. Thanks very much!
     
    Last edited: Oct 19, 2012
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