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

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.
  2. jcsd
  3. Oct 18, 2012 #2


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook