Let \(A\) be the following matrix (over \(\mathbb{Q}\)) \[\begin{pmatrix} -2 & 1 & 1 \\ -4 & 3 & 1 \\ -6 & 6 & -1 \end{pmatrix}.\] Find the rational canonical form for \(A\). Give a \(3 \times 3\) matrix \(B\) which has the same characteristic polynomial as \(A\), but which is not similar to \(A\).
Solution
The rational canonical form and a suitable second matrix are \[ C=\begin{pmatrix}0&0&2\\1&0&3\\0&1&0\end{pmatrix}, \qquad B=\begin{pmatrix}2&0&0\\0&-1&0\\0&0&-1\end{pmatrix}. \]
1. The vector \(v=(1,0,0)^{\mathsf T}\) is cyclic for \(A\),and \(P^{-1}AP=C\) for \(P=(v\ Av\ A^2v)\).
Direct multiplication gives \[ Av=\begin{pmatrix}-2\\-4\\-6\end{pmatrix},\quad A^2v=\begin{pmatrix}-6\\-10\\-6\end{pmatrix},\quad A^3v=\begin{pmatrix}-4\\-12\\-18\end{pmatrix}=2v+3Av. \] Thus \[ P=(v\ Av\ A^2v) =\begin{pmatrix}1&-2&-6\\0&-4&-10\\0&-6&-6\end{pmatrix}, \qquad \det P=24-60=-36\ne0. \] Its columns form a basis. In this basis, \(A\) sends the first vector to the second, the second to the third, and the third to twice the first plus three times the second. These three identities say exactly that \(AP=PC\), so \(P^{-1}AP=C\).
2. The matrix \(C\) is the rational canonical form, withsingle invariant factor\(f(t)=t^3-3t-2=(t-2)(t+1)^2\).
The companion matrix of a monic polynomial \(t^3+c_2t^2+c_1t+c_0\) has subdiagonal entries one and last column \((-c_0,-c_1,-c_2)^{\mathsf T}\) [@DF04]. Hence \(C\) is the companion matrix of \(f\). The cyclic basis in step 1 identifies the \(\mathbb Q[t]\)-module with \(\mathbb Q[t]/(f)\): the map \(g\mapsto g(A)v\) is onto; division by \(f\) reduces any element in its kernel to a polynomial of degree at most two, whose coefficients must vanish by independence of \(v,Av,A^2v\). Here \(f(A)v=0\) is the displayed identity for \(A^3v\). Thus the kernel is exactly \((f)\), establishing the single invariant factor and the asserted rational form.
Expanding the determinant of \(tI-C\) gives \(t(t^2-3)-2=t^3-3t-2\). Similarity therefore gives \(\chi_A=f\). The cyclic-basis argument also proves that the minimal polynomial has degree three and equals \(f\).
3. The matrix \(B\) has the same characteristic polynomialas \(A\) but is not similar to \(A\).
Its diagonal entries give \(\chi_B(t)=(t-2)(t+1)^2=\chi_A(t)\). On the other hand, \[ A+I=\begin{pmatrix}-1&1&1\\-4&4&1\\-6&6&0\end{pmatrix} \] has rank two. Its second column is the negative of the first, so its rank is at most two, while the minor in rows one and two and columns one and three is \((-1)(1)-(1)(-4)=3\ne0\). But \(B+I=\operatorname{diag}(3,0,0)\) has rank one. If \(A\) and \(B\) were similar, adding the identity would preserve that similarity and hence their ranks, a contradiction.