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Thread: about diagonalization

  1. #1
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    about diagonalization

    Hello,

    I want to write the following the matrix
    0 $\displaystyle \beta $ 0
    1 0 0
    0 0 0

    in the form $\displaystyle C^{-1}AC$


    over $\displaystyle Mat(3,\mathbb{Q}(\sqrt{\beta}))$ and A is a diagonal matrix. I know that $\displaystyle \sqrt{\beta}\notin \mathbb{Q}$.


    If there is a method to solve that kind of problem I would apperciate your help.

    Thanks in advance.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Biscaim View Post
    Hello,

    I want to write the following the matrix
    0 $\displaystyle \beta $ 0
    1 0 0
    0 0 0

    in the form $\displaystyle C^{-1}AC$


    over $\displaystyle Mat(3,\mathbb{Q}(\sqrt{\beta}))$ and A is a diagonal matrix. I know that $\displaystyle \sqrt{\beta}\notin \mathbb{Q}$.


    If there is a method to solve that kind of problem I would apperciate your help.

    Thanks in advance.
    we really don't need the condition $\displaystyle \sqrt{\beta} \notin \mathbb{Q}.$ what we need for our matrix to be diagonalizable is $\displaystyle \beta \neq 0.$ let $\displaystyle B=\begin{pmatrix}0 & \beta & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$ the eigenvalues of $\displaystyle B$ are $\displaystyle 0, \pm \sqrt{\beta}.$ thus: $\displaystyle A=\begin{pmatrix}0 & 0 & 0 \\ 0 & \sqrt{\beta} & 0 \\ 0 & 0 & -\sqrt{\beta} \end{pmatrix}.$

    then find the eigenvectors corresponding to each eigenvalue and put them in a matrix so that the first column is the eigenvector corresponding to the first eigenvalue on $\displaystyle A,$ i.e. $\displaystyle 0$ and the

    second and the third columns are the eigenvectors corresponding to $\displaystyle \sqrt{\beta}$ and $\displaystyle -\sqrt{\beta}$ respectively. the result would be this matrix: $\displaystyle X=\begin{pmatrix}0 & \sqrt{\beta} & -\sqrt{\beta} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix}.$ this matrix satisfies $\displaystyle B=XAX^{-1}.$

    the matrix $\displaystyle C$ that you're looking for is just the inverse of $\displaystyle X.$ you'll get: $\displaystyle C=X^{-1}=\begin{pmatrix}0 & 0 & 1 \\ \frac{1}{2\sqrt{\beta}} & \frac{1}{2} & 0 \\ -\frac{1}{2\sqrt{\beta}} & \frac{1}{2} & 0 \end{pmatrix}.$ [note that the entries of all matrices are in $\displaystyle \mathbb{Q}(\sqrt{\beta}).$]
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