Results 1 to 9 of 9

Thread: Matrix element of a field

  1. #1
    Junior Member
    Joined
    Sep 2011
    Posts
    34

    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.

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

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

    The characteristic polynomial of A:

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

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

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

    to solve for x and y for both eigenvalues.

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

    Am I correct in my working out so far?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Matrix element of a field

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

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

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

    so, the eigenvalues of $\displaystyle A$ are $\displaystyle 0+i,0-i$ or equivalently $\displaystyle i,2i$ .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2011
    Posts
    34

    Re: Matrix element of a field

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

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

    so, the eigenvalues of $\displaystyle A$ are $\displaystyle 0+i,0-i$ or equivalently $\displaystyle i,2i$ .
    Oh yes, 3 isn't an element of $\displaystyle F_3$, 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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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 $\displaystyle F_3$ we have $\displaystyle 1+2=0$ so, $\displaystyle 2=-1$ .

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

    As $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2011
    Posts
    34

    Re: Matrix element of a field

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

    As $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2011
    Posts
    34

    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 $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .
    How does $\displaystyle \dim \ker(A-iI)=1$ and how do I go from there to find the eigenvectors?

    Thanks again for your help.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Matrix element of a field

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

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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:

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

    that is:

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

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

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Jun 16th 2011, 06:30 AM
  2. Derivative of matrix element
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Feb 27th 2010, 09:05 AM
  3. Algebra over a field - zero element
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Sep 9th 2009, 05:05 AM
  4. Exponention in Composite Field Element
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Aug 15th 2009, 02:06 AM
  5. Element in finite field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 19th 2009, 04:33 PM

Search Tags


/mathhelpforum @mathhelpforum