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Math Help - Matrix element of a field

  1. #1
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    Matrix element of a field

    Hi, I am not 100% sure my working is correct for this question. Could someone be kind enough to check my working please.

    F= \{a+bi | a,b \in F_3 \}

    A=\dbinom{2 \ \ 1}{1 \ \ 1} \in F^{2x2}

    The characteristic polynomial of A:

    det \dbinom{2-\lambda \ \ \ \ 1}{1 \ \ \ \ \ 1-\lambda}= \lambda -3\lambda +1

    So the eigenvalues are \lambda_1 = \frac{-(\sqrt{5}-3)}{2} and \lambda_2 = \frac{\sqrt{5}+3}{2}.

    To find the eigenvectors we sub in \lambda_1 and \lambda_2 into the equations:
    (2-\lambda)x+y=0
    x+(1-\lambda)y=0

    to solve for x and y for both eigenvalues.

    So for example, for \lambda_2 I get y=\frac{3(\sqrt{5}+2)}{2} and x=\frac{3(6\sqrt{5}+13)}{4}.

    Am I correct in my working out so far?

    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Matrix element of a field

    Quote Originally Posted by Juneu436 View Post
    F= \{a+bi | a,b \in F_3 \} A=\dbinom{2 \ \ 1}{1 \ \ 1} \in F^{2x2} The characteristic polynomial of A: det \dbinom{2-\lambda \ \ \ \ 1}{1 \ \ \ \ \ 1-\lambda}= \lambda -3\lambda +1 So the eigenvalues are \lambda_1 = \frac{-(\sqrt{5}-3)}{2} and \lambda_2 = \frac{\sqrt{5}+3}{2}.

    I suppose F_3=\mathbb{Z}/(3)=\{0,1,2\} . Then, the characteristic polynomial of A is

    \chi(\lambda)=\lambda ^2-(2+1)\lambda +1=\lambda^2-0\lambda +1=\lambda^2+1

    so, the eigenvalues of A are 0+i,0-i or equivalently i,2i .
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  3. #3
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    Re: Matrix element of a field

    Quote Originally Posted by FernandoRevilla View Post
    I suppose F_3=\mathbb{Z}/(3)=\{0,1,2\} . Then, the characteristic polynomial of A is

    \chi(\lambda)=\lambda ^2-(2+1)\lambda +1=\lambda^2-0\lambda +1=\lambda^2+1

    so, the eigenvalues of A are 0+i,0-i or equivalently i,2i .
    Oh yes, 3 isn't an element of F_3, thanks for that.

    Why is -i equivalent to 2i? I know it is a stupid question.
    Then how do I work out the eigenvectors?

    Thanks
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Matrix element of a field

    Quote Originally Posted by Juneu436 View Post
    Why is -i equivalent to 2i? I know it is a stupid question.
    Because in F_3 we have 1+2=0 so, 2=-1 .

    Then how do I work out the eigenvectors?
    For example, if \lambda=i then \ker (A-iI)\equiv \begin{Bmatrix} (2-i)x_1+x_2=0\\x_1+(1-i)x_2=0\end{matrix}

    As i is simple, \dim \ker (A-iI)=1 . Solving you'll find a basis, for example B_i=\{(1,1+i)\} .
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  5. #5
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    Re: Matrix element of a field

    Quote Originally Posted by FernandoRevilla View Post
    For example, if \lambda=i then \ker (A-iI)\equiv \begin{Bmatrix} (2-i)x_1+x_2=0\\x_1+(1-i)x_2=0\end{matrix}

    As i is simple, \dim \ker (A-iI)=1 . Solving you'll find a basis, for example B_i=\{(1,1+i)\} .
    Thanks FernandoRevilla, but I don't really understand this. Could you please explain?
    I can't see how to get the eigenvectors.

    Thanks, and sorry for being a pain.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Matrix element of a field

    Quote Originally Posted by Juneu436 View Post
    Thanks FernandoRevilla, but I don't really understand this. Could you please explain? I can't see how to get the eigenvectors.
    Could you specify your doubts?. These things are just routine knowing the corresponding theory.
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  7. #7
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    Re: Matrix element of a field

    Quote Originally Posted by FernandoRevilla View Post
    Could you specify your doubts?. These things are just routine knowing the corresponding theory.
    This section:
    As i is simple, \dim \ker (A-iI)=1 . Solving you'll find a basis, for example B_i=\{(1,1+i)\} .
    How does \dim \ker(A-iI)=1 and how do I go from there to find the eigenvectors?

    Thanks again for your help.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Matrix element of a field

    Quote Originally Posted by Juneu436 View Post
    How does \dim \ker(A-iI)=1 and how do I go from there to find the eigenvectors?
    There is a well known theorem: if \lambda is a simple eigenvalue then, the corresponding eigenspace V_{\lambda} has dimension 1. This means that a non null vector of V_{\lambda} form a basis of V_{\lambda} (because is linearly independent). In our case, and choosing (for example) x_1=1 we obtain from the first equation that x_2=1+i . You needn't verify the second equation because necessarily the rank of the system matrix is 1.
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  9. #9
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    Re: Matrix element of a field

    or you could just solve the system Av = λv for each eigenvalue λ.

    for example, with λ = i, and writing a+bi = z, c+di = w this leads to:

    \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}z\\w\end{bmatrix} = \begin{bmatrix}iz\\iw\end{bmatrix}

    that is:

    \begin{bmatrix}2z+w\\z+w \end{bmatrix}  = \begin{bmatrix}iz\\iw\end{bmatrix} so:

    z = iz - iw
    z(1+2i) = 2iw
    z(1+2i)(i) = -2w = w
    w = (-2+i)z = (1+i)z

    so any eigenvector corresponding to i is of the form (z,(1+i)z)...z, of course, must be non-zero. it is convenient to choose z = 1+0i = 1.
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