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$\textbf{Exercise.}$ Let be $\theta = \sqrt[4]{2}$, $V = \mathbb{Q}(\theta)$. Let be $f: V \longrightarrow V$ a linear transformation over $\mathbb{Q}$ defined by $f(v) := \theta v$. Show that the matrix of $f$ on the basis $\mathcal{B} = \{ 1, \theta, \theta^2, \theta^3 \}$ is

$$A = \begin{pmatrix} 0 & 0 & 0 & 2\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}$$

Find the Canonical Jordan Form of the matrix $A$.

I'm trying to remember how to compute the Canonical Jordan Form. I would like to know if my attempt it's correct and if there is an easier way than this to compute the Jordan Form.

$\textbf{My attempt:}$

Computing the caractheristic polynomial, I found $p_A(x) = \det \left( xId - A \right) = x^4 - 2$, then the eigenvalues are $\lambda_1 = - \theta$ and $\lambda_2 = \theta$.

I computed the eigenspaces associated to $\lambda_1$ and $\lambda_2$, which I will denote by $E(\lambda_1)$ and $E(\lambda_2)$ respectively, and I found

$E(\lambda_1) = \text{Ker} (A + \theta Id) = \left[ (-\theta^3,\theta^2,-\theta,1) \right]$ and $E(\lambda_2) = \text{Ker} (A - \theta Id) = \left[ (\theta^3,\theta^2,\theta,1) \right]$,

but I needed two more vectors in order to find the matrix $P$ such that $J = P^{-1}AP$. I observed that

$\text{Ker} \left( (A + \theta Id)^2 \right) = \left[ (3 \theta^2,-2 \theta, 1,0), (2\theta^3,-\theta^2,0,1) \right]$

and

$\text{Ker} \left( (A - \theta Id)^2 \right) = \left[ (3 \theta^2, 2 \theta, 1,0), (-2\theta^3,-\theta^2,0,1) \right],$

then I choose two eigenvectors $v_1 = (3 \theta^2,-2 \theta, 1,0)$ and $v_2 = (3 \theta^2,2 \theta, 1,0)$ and I constructed the matrix $P$ using this eigenvectors and the generators of $E(\lambda_1)$ and $E(\lambda_2)$:

$P = \begin{bmatrix} - \theta^3 & 3 \theta^2 & \theta^3 & 3\theta^2\\ \theta^2 & -2\theta & \theta^2 & 2\theta\\ -\theta & 1 & \theta & 1\\ 1 & 0 & 1 & 0 \end{bmatrix}$

I'm stuck in compute $P^{-1}$, because it's too much working. I would like to know if my attempt is correct until now and if there is an easier way than this to find the Jordan Form.

Thanks in advance!

$\textbf{EDIT:}$

I forgot to say, but I tried compute the Canonical Jordan's form, but I'm stuck too, I think it's the same reason in this topic, i.e., I know how compute the Jordan's form when the eigenvalues are real, but I don't know how to proceed when I have complex eigenvalues. Reading the topic previously quoted, I tried again compute the Jordan's form:

Since $\dim_{\mathbb{Q}} E(\lambda_1) = 1$, $\dim_{\mathbb{Q}} E(\lambda_2) = 1$ and $p_A(x) = (x - \theta) (x + \theta) (x^2 + \theta^2)$, we have

$$J = \begin{bmatrix} -\theta & 0 & 0 & 0\\ 0 & \theta & 0 & 1\\ 0 & 0 & 0 & 2\\ 0 & 0 & -2 & 0 \end{bmatrix},$$

where the Jordan's block with a complex root $\alpha + i \beta$ has a general form

enter image description here

according the lecture notes which I'm studying.

I have some questions about my second attempt:

  1. Is it correct?

  2. Has difference considering $1$ on entries of the Jordan's block with complex roots? Because here the author of the question doesn't consider $1$ on entries of the Jordan's block.

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    $\begingroup$ Typically, "find the Jordan Canonical form" means that they want you to find $J$, but not necessarily $P$. That should make this problem much easier $\endgroup$ Sep 28, 2018 at 17:23

1 Answer 1

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The decomposition field of $A$ is $K=\mathbb{Q}(\theta,i)\subset \mathbb{C}$; then the standard Jordan form (JF) is over $K$; since the eigenvalues are simple, $A$ is diagonalizable and its JF is $diag(-\theta,\theta,i\theta,-i\theta)$.

Now, there exists an extended Jordan form over $\mathbb{Q}(\theta)\subset\mathbb{R}$. This is $diag(-\theta,\theta,\begin{pmatrix}0&\theta\\-\theta&0\end{pmatrix})$

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