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Eigenvalues of A-adjoint and A

  1. Jan 29, 2012 #1
    Eigenvalues of A* and A

    Show that the eigenvalues of A* are conjugates of the eigenvalues of A.

    I know this is an easy problem, but I've just been spinning my wheels manipulating the equations with the transpose, conjugate, and adjoint properties.

    [itex]
    \begin{align}

    A^* = \bar{A}^T\\

    A\vec{x} = \lambda\vec{x}\\

    A^*\vec{x} = \bar\lambda\vec{x}\\
    \end{align}

    [/itex]
     
    Last edited: Jan 29, 2012
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  3. Jan 29, 2012 #2

    Dick

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    [itex]\lambda[/itex] is an eigenvalue of [itex]A[/itex] if [itex] det(A-\lambda I)=0[/itex]. Try taking conjugate and transpose on that.
     
  4. Jan 29, 2012 #3
    You know, I thought about doing it like that but didn't because I figured the other way would be easier.

    [itex]
    \begin{align}
    det(A-\lambda I) = 0\\\\
    \overline{det(A-\lambda I)} = \bar0\\
    det(\overline{A-\lambda I)} = 0\\
    = det(\bar A-\bar\lambda I) = 0\\\\
    = det(\bar A-\bar\lambda I)^T = 0^T\\
    = det(\bar A^T-\bar\lambda I) = 0\\
    = det(\bar A^T-\bar\lambda I) = 0\\
    = det(A^*-\bar\lambda I)=0\\
    \\

    \end{align}
    [/itex]
     
    Last edited: Jan 29, 2012
  5. Jan 29, 2012 #4

    Dick

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    Just to make sure you are clear on this. [itex]A-\lambda I[/itex] is NOT EQUAL to [itex]A^* - \bar{\lambda} I[/itex]. What is equal is [itex]det(A-\lambda I)=det(A^* - \bar{\lambda} I)[/itex].
     
  6. Jan 29, 2012 #5
    Yes, I know that.

    The characteristic polynomials for A and A* are equal?
     
    Last edited: Jan 29, 2012
  7. Jan 29, 2012 #6

    Dick

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    Well, no. That's not what they are asking you to show. If A=[[1,0],[0,i]], then how does the characteristic polynomial of A compare with the characteristic polynomial of A*?
     
  8. Jan 29, 2012 #7
    I know what they're asking me. You're right. The characteristic polynomials aren't equal. I was overlooking a small detail.

    The two characteristic polynomials are complex conjugates, right?

    I'm not quite sure why this is true: [itex]det(A-\lambda I)=det(A^* - \bar{\lambda} I)[/itex].
     
    Last edited: Jan 29, 2012
  9. Jan 29, 2012 #8
    Those two determinants are equal because they are both zero.
     
  10. Jan 29, 2012 #9

    Dick

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    Right. Since they are complex conjugates of each other, if one equals 0 then the other equals zero.
     
  11. Jan 29, 2012 #10
    Oh, okay. I didn't know the characteristic polynomials were complex conjugates of each other.

    I think I have it written out more correctly now in post #3.
     
  12. Jan 29, 2012 #11

    Dick

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    Post 3 still looks a little funny. The properties of determinant that you need are [itex]det(\bar M)=\overline{det(M)}[/itex] and [itex]det(M^T)=det(M)[/itex].
     
  13. Jan 29, 2012 #12
    Is it okay now?
     
  14. Jan 29, 2012 #13

    Dick

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    Okay. I wouldn't bother with transposing a scalar. Notice if you start with [itex]det(A-\lambda I)=c[/itex] then you wind up with [itex]det(A^*-\bar{\lambda} I)=\bar c[/itex]
     
  15. Jan 29, 2012 #14
    Oh, you're right. I mistakenly treated it as a vector.

    Thanks for the help.
     
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