Let \(A\) be the real matrix \[ A = \begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 3 \\ -3 & -3 & -5 \end{bmatrix}. \] Find an invertible \(3 \times 3\) real matrix \(P\) such that \(P^{-1}AP\) is in rational form.
Solution
One valid choice is \[ \boxed{ P=\begin{pmatrix} 1&1&1\\ 0&0&3\\ -1&0&-3 \end{pmatrix}, \qquad P^{-1}AP= \begin{pmatrix} -2&0&0\\ 0&0&2\\ 0&1&-1 \end{pmatrix}.} \] The lower-right block is the companion matrix of \(x^2+x-2=(x+2)(x-1)\) in Hoffman–Kunze’s convention.
1. Determine the invariant factors.
A direct determinant calculation gives \[ \chi_A(x)=\det(xI-A)=(x-1)(x+2)^2. \] Also \[ A+2I= \begin{pmatrix} 3&3&3\\3&3&3\\-3&-3&-3 \end{pmatrix} \] has rank one, so the \(-2\) eigenspace has dimension two. The \(1\) eigenspace is one-dimensional, for example spanned by \((1,1,-1)\). Thus \(A\) is diagonalizable and its minimal polynomial is \[ m_A(x)=(x+2)(x-1)=x^2+x-2. \] The invariant factors multiply to the characteristic polynomial and the last one is the minimal polynomial [@HK71]. Hence they are \[ x+2, \qquad (x+2)(x-1). \]
2. Choose cyclic generators for the two invariant-factor summands.
Let \[ u=(1,0,-1), \qquad v=(1,0,0). \] Then \[ Au=-2u, \] so \(Z(u;A)=\operatorname{span}\{u\}\) has annihilator \(x+2\). Moreover \[ Av=(1,3,-3). \] The three vectors \[ u,\quad v,\quad Av \] form a basis because the matrix with these columns is \[ P= \begin{pmatrix} 1&1&1\\ 0&0&3\\ -1&0&-3 \end{pmatrix} \] and \(\det P=-3\ne0\). Thus \(V=Z(u;A)\oplus Z(v;A)\), with \(Z(v;A)\) two-dimensional.
Since \(m_A(A)=0\), \[ (A^2+A-2I)v=0, \] so \[ A(Av)=2v-Av. \] Thus relative to the ordered cyclic basis \((v,Av)\), the restriction of \(A\) is \[ \begin{pmatrix}0&2\\1&-1\end{pmatrix}, \] the companion matrix of \(x^2+x-2\) [@HK71].
3. Assemble the rational form.
Relative to the ordered basis \((u,v,Av)\), the first summand is multiplication by \(-2\) and the second has the companion matrix computed above. Therefore \[ P^{-1}AP =\operatorname{diag}\left(-2, \begin{pmatrix}0&2\\1&-1\end{pmatrix}\right) = \begin{pmatrix} -2&0&0\\ 0&0&2\\ 0&1&-1 \end{pmatrix}, \] which is the rational canonical form corresponding to the invariant factors \(x+2\mid(x+2)(x-1)\).