Find the rational canonical form of \[ \begin{pmatrix}2&3\\7&1\end{pmatrix}. \]
Find all possible rational canonical forms for 4-by-4 matrices over the complex numbers whose characteristic polynomial is \((x^2+1)(x+1)^2\).
Solution
For a monic polynomial \(f=x^d+c_{d-1}x^{d-1}+\cdots+c_0\), write \(C(f)\) for its companion matrix with ones on the subdiagonal and last column \((-c_0,\ldots,-c_{d-1})^{\mathsf T}\).
1. The answer to part (a) is\[ C(x^2-3x-19)=\begin{pmatrix}0&19\\1&3\end{pmatrix}. \]
Let \(A\) be the given matrix. Its characteristic polynomial is \[ \det(xI-A)=(x-2)(x-1)-21=x^2-3x-19. \] For \(v=(1,0)^{\mathsf T}\), the columns \(v,Av\) form the matrix \[ P=\begin{pmatrix}1&2\\0&7\end{pmatrix},\qquad \det P=7\ne0. \] Writing \(C=\left(\begin{smallmatrix}0&19\\1&3\end{smallmatrix}\right)\), direct multiplication gives \[ AP=PC=\begin{pmatrix}2&25\\7&21\end{pmatrix}. \] Thus \(P^{-1}AP=C\). The cyclic basis \(v,Av\) gives the single invariant factor \(x^2-3x-19\), so this companion matrix is the rational canonical form over \(\mathbb Q\) [@DF04].
2. There are exactly two invariant-factor chains in part (b):\[ \bigl((x^2+1)(x+1)^2\bigr),\qquad \bigl(x+1,\ (x^2+1)(x+1)\bigr). \]
Over \(\mathbb C\), the characteristic polynomial is \((x-i)(x+i)(x+1)^2\). Let \(f_1\mid\cdots\mid f_r\) be the nonconstant monic invariant factors. Their product is the characteristic polynomial [@DF04]. If \(x-i\) or \(x+i\) divided some \(f_j\) with \(j<r\), it would also divide \(f_r\), contradicting its multiplicity one in the product. Hence both of those factors occur only in \(f_r\).
Every earlier \(f_j\) is consequently a positive power of \(x+1\). If there is an earlier factor, divisibility forces \(x+1\) into \(f_r\) as well. Its total multiplicity is only two, so \(r=2\), \(f_1=x+1\), and \(f_2=(x^2+1)(x+1)\). If there is no earlier factor, \(r=1\) and \(f_1\) is the whole characteristic polynomial. This proves that the two chains exhaust all possibilities.
3. The two rational canonical forms are\[ \begin{pmatrix} 0&0&0&-1\\ 1&0&0&-2\\ 0&1&0&-2\\ 0&0&1&-2 \end{pmatrix},\qquad \begin{pmatrix} -1&0&0&0\\ 0&0&0&-1\\ 0&1&0&-1\\ 0&0&1&-1 \end{pmatrix}. \]
The products expand to \[ (x^2+1)(x+1)^2=x^4+2x^3+2x^2+2x+1, \qquad (x^2+1)(x+1)=x^3+x^2+x+1. \] The first displayed matrix is the companion matrix of the quartic. The second is \(C(x+1)\oplus C(x^3+x^2+x+1)\). Their block polynomials form the two divisibility chains in step 2, so both are rational canonical forms with the required characteristic polynomial. Their minimal polynomials are their largest invariant factors, of degrees four and three respectively [@DF04]. They are therefore not similar, and step 2 proves completeness.