I'm given the matrix $ A = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 3 & 3 \\ 2 & 2 & 4 \end{pmatrix}$ over $ \mathbb{Q} $.
I need to prove that $ \mathbb{Q}^3 $ is an A cyclic space (meaning, a cyclic module over $ Q[x] $) and to explicitly find a spanning vector.
From the theory i know that i can find invertible matrices, $ P $ and $ Q $ such that $ P (\lambda I - A) Q = B $ and that $ B $ is diagonal.
Indeed after eliminating the matrix, i get that there are some $ P, Q $ such that:
$ P (\lambda I - A) Q = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \lambda (\lambda^2-6\lambda+7) \end{pmatrix} $
Now i can show that $ \mathbb{Q}^3 \cong \frac{\mathbb{Q}[x]}{<\lambda (\lambda^2-6\lambda+7)>} \cong \mathbb{Q[x]}z $.
For explicitly computing z, i have to actually compute $ Q $, $ P $, and $ Q^{-1} $. Is there any way of explicitly computing z without that much of calculations?
Yes; randomly choose $z\in \mathbb{Q}^3$ and test it!
For example $z=[1,7,12]^T$ or $z=[2,5,9]^T$ are convenient.
EDIT. If you randomly choose $z$ in $\mathbb{R}^3$ (using iid normal laws for each component) then $prob(z \text{ is a cyclic vector})=1$. If you choose the $z_i$ uniformly in $[[-1000..1000]]$, then a random test shows that $prob(z \text{ is a cyclic vector})\approx 1-4\times 10^{-4}$.
To avoid any doubt, just check if $\det(z,Az,A^2z)\not= 0$.