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Finding eigenvalues of matrix with complex entries

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the eigenvalues/eigenvectors of the unitary matrix

    A = (1/√5){{1,2},{2i,-i}}

    2. Relevant equations

    det[A-λI]=0

    AA* = I (where A* denotes the adjoint of A)


    3. The attempt at a solution

    I have tried to do this straightforwardly, as I would any other matrix, by using the equation above. However, I end up with the following quadratic equation:

    λ^2+λi-λ-5i = 0

    My first thought was to group this according to real and imaginary parts and then set both to zero:

    λ^2 - λ + (λ-5)i = 0

    But this actually gives me 3 eigenvalues and none of them is correct, anyway. So that's a dead end (although I don't understand why; the logic makes sense to me: if a complex number is zero, then it's real part must be zero and it's imaginary part must be zero. Where did I go wrong here? Maybe in not accounting for the possibility of complex eigenvalues, so my grouping is not correct?)

    My next tactic was to put it into the classic form of a quadratic equation, like so:

    λ^2+(i-1)λ-5i=0

    and then apply the good old quadratic equation with a = 1, b = (i-1), and c = -5i. This gives me

    (1/2)(-i+1±3*Sqrt(2i))

    This is not only not the correct answer, but I don't even know how to make sense of it. (What is the square root of i? Just the 1/4th power of -1, I suppose?) Can anyone offer some guidance as to where I've gone wrong here?

    Am I perhaps supposed to be using the fact that the matrix is unitary?
     
  2. jcsd
  3. Nov 18, 2012 #2

    I like Serena

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    Hi mindarson! :smile:

    Your straightforward method is fine.
    However, your quadratic equation is incorrect.
    Did you factor the (1/√5) into the matrix before you tried to find the equation?
    If not, could you do so?


    Your method to solve the quadratic equation is also fine.
    To answer your question how to draw the square root of i, are you aware of the polar representation of an imaginary number?
    That is:
    $$z=r e^{i \phi}$$
    In particular:
    $$i=e^{i \frac \pi 2}$$
    This means that the square root of i is:
    $$\sqrt i = \sqrt{e^{i \frac \pi 2}} = (e^{i \frac \pi 2})^{\frac 1 2} = e^{i \frac \pi 4} = \frac 1 {\sqrt{2}}(1 + i)$$
     
  4. Nov 18, 2012 #3

    micromass

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    You really shouldn't use the square root symbol on imaginary numbers. It's very bad form. And "the" square root of i doesn't exist. There are two square roots though, but I don't think we should call it "the" square root. You can talk about the prinicipal root though.
     
  5. Nov 18, 2012 #4
    I actually did try this in my very first attempt. However, I didn't think it would matter, since it would just factor out of the characteristic polynomial and then when I divided both sides of the quadratic equation by it, it would just vanish anyway. I will give it another try, though.

    Thanks for the insight into how to represent 'the' square root of i, or however proper math folks refer to it. :)

    When you say that my method for solving the quadratic equation is fine, do you mean using the quadratic formula as I outlined?
     
  6. Nov 18, 2012 #5
    Okay, then, what is the proper way to refer to this (these) object(s) that showed up in my calculation? Would it be, say, i^(1/2) or something like that?
     
  7. Nov 18, 2012 #6

    micromass

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    The notation [itex]i^{\frac{1}{2}}[/itex] is used a lot, but you got to be careful. Sometimes it is used to denote multiple values (since exponentiation is multiple valued). Sometimes it denotes an (arbitrarily chosen) principal value. In that case, you should say how you choose that value.

    I personally would say: take an [itex]\alpha\in \mathbb{C}[/itex] such that [itex]\alpha^2=i[/itex]. But again: there are two such [itex]\alpha[/itex].
     
  8. Nov 18, 2012 #7
    I just tried this, and it is not yielding the correct eigenvalues. I end up with the quadratic equation

    λ^2 + ((i-1)/√5)λ -i = 0

    If I solve this using quadratic formula, I get incorrect eigenvalues.
     
  9. Nov 18, 2012 #8
    Perhaps I should mention that what I'm ultimately trying to do here is to diagonalize the given matrix. That is why I'm trying to find the eigenvalues. Is there maybe some other way to diagonalize a unitary matrix? Because this just isn't working, and it feels wrong somehow. I've diagonalized many matrices before, but they have always been real or Hermitian. Trying to diagonalize this one is just frustrating me, because I'm getting eigenvalues that don't make sense (to me) and are clearly wrong.

    Anyhow, thanks for the help.
     
  10. Nov 19, 2012 #9

    I like Serena

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    Yes.

    As mm said, in general you have to be careful with square roots and broken powers of imaginary numbers.
    However, in this particular case (solving a quadratic equation) it does not matter.
    The reason is that the square root already has a ± in front of it.
    This takes care of the 2 solutions mm is referring to (just like it does in the case of real numbers).


    This is the correct quadratic equation.
    Good!

    So how did you solve it?

    Again, your method is fine.
    To diagonalize a matrix the typical first step is to find the eigenvalues.
     
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