Let the matrix
\begin{equation}
A=\begin{bmatrix}
2 & 1 & 2 \\
-2 & -1 & -4 \\\
1 & 1 & 3
\end{bmatrix}.
\end{equation}
So far I found the characteristic polynomial $C_A(x)=(2-x)(x-1)^2$, the minimal polynomial $m_A(x)=(2-x)(x-1)$ and the Jordan matrix form is
\begin{equation}
J=\begin{bmatrix}
2 & * & *\\
* & 1 & * \\\
* & * & 1
\end{bmatrix}.
\end{equation}
Does someone could tell me from what I did how it possible to find the rational canonical form of $A$?
Best Answer
There are two "rational canonical forms", one obtained using elementary divisors and another obtained by invariant factors.
The invariant factors form a list where each term divides the next, their product is the characteristic polynomial, and the minimal polynomial is the largest one.
Therefore in this problem, $(x-2)(x-1)$ is the largest invariant factor. The other invariant factor is $(x-1)$ simply because there is nothing left over.
The rational canonical form is obtained by putting the companion matrices corresponding to the invariant factors on the main diagonal.
So the rational canonical form of $A$ given by the invariant factors is: $$\begin{pmatrix} 0 & -2 & 0\\ 1 & 3 & 0\\ 0 & 0 & 1 \end{pmatrix}_.$$
The matrix $$\begin{pmatrix} 0 & -2 \\ 1 & 3 \end{pmatrix}_.$$
Is the companion matrix of $(x-2)(x-1)$.
https://en.wikipedia.org/wiki/Companion_matrix