Find the rational canonical form over \(\mathbb{Q}\) of the following matrix:
\[\begin{pmatrix} 3 & -1 & 10 \\ 0 & 2 & 5 \\ 0 & 0 & 3 \end{pmatrix}.\]
Solution
Let \(A\) denote the given matrix. Its rational canonical form is \[ C=\begin{pmatrix} 0&0&18\\ 1&0&-21\\ 0&1&8 \end{pmatrix}, \] with the single invariant factor \(f(T)=(T-3)^2(T-2)=T^3-8T^2+21T-18\). We use the companion-matrix convention with ones on the subdiagonal.
1. The vector \(v=(0,0,1)^t\) is cyclic for \(A\).
Because \(A\) is upper triangular, \[ \det(TI-A)=(T-3)^2(T-2)=f(T). \] Direct multiplication gives \[ Av=\begin{pmatrix}10\\5\\3\end{pmatrix},\qquad A^2v=\begin{pmatrix}55\\25\\9\end{pmatrix}. \] The matrix with columns \(v,Av,A^2v\) is \[ P=\begin{pmatrix}0&10&55\\0&5&25\\1&3&9\end{pmatrix}, \qquad \det P=10\cdot25-55\cdot5=-25\ne0. \] Thus these three vectors form a basis of \(\mathbb Q^3\).
2. In that basis, \(A\) has the displayed companion matrix \(C\).
The exact matrix products are \[ AP=PC= \begin{pmatrix} 10&55&230\\ 5&25&95\\ 3&9&27 \end{pmatrix}. \] Since \(P\) is invertible, this proves \(P^{-1}AP=C\). Equivalently, the last column states \(A^3v=18v-21Av+8A^2v\), while the first two columns advance along the cyclic basis. A cyclic module with this degree-three relation has the single invariant factor \(f\), and its rational canonical form is the companion matrix \(C\) [@DF04].