Hello,

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

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

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

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

2. Originally Posted by Biscaim
Hello,

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

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

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

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

we really don't need the condition $\sqrt{\beta} \notin \mathbb{Q}.$ what we need for our matrix to be diagonalizable is $\beta \neq 0.$ let $B=\begin{pmatrix}0 & \beta & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$ the eigenvalues of $B$ are $0, \pm \sqrt{\beta}.$ thus: $A=\begin{pmatrix}0 & 0 & 0 \\ 0 & \sqrt{\beta} & 0 \\ 0 & 0 & -\sqrt{\beta} \end{pmatrix}.$
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 $A,$ i.e. $0$ and the
second and the third columns are the eigenvectors corresponding to $\sqrt{\beta}$ and $-\sqrt{\beta}$ respectively. the result would be this matrix: $X=\begin{pmatrix}0 & \sqrt{\beta} & -\sqrt{\beta} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix}.$ this matrix satisfies $B=XAX^{-1}.$
the matrix $C$ that you're looking for is just the inverse of $X.$ you'll get: $C=X^{-1}=\begin{pmatrix}0 & 0 & 1 \\ \frac{1}{2\sqrt{\beta}} & \frac{1}{2} & 0 \\ -\frac{1}{2\sqrt{\beta}} & \frac{1}{2} & 0 \end{pmatrix}.$ [note that the entries of all matrices are in $\mathbb{Q}(\sqrt{\beta}).$]