# Computing Canonical Jordan Form over a field $\mathbb{Q}$

$$\textbf{Exercise.}$$ 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.

$$\textbf{EDIT:}$$

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

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.

• 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 Sep 28, 2018 at 17:23

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 $$diag(-\theta,\theta,\begin{pmatrix}0&\theta\\-\theta&0\end{pmatrix})$$