Find a Rational Canonical Form and a Jordan Canonical Form for the following \(3 \times 3\) matrix with real entries: \[A = \begin{pmatrix} 0 & -1 & 2 \\ 3 & -4 & 6 \\ 2 & -2 & 3 \end{pmatrix}.\]
Describe all distinct isomorphism classes of Abelian groups of order 360.
Solution
1. The characteristic polynomial of \(A\) is \(p_A(\lambda)=(\lambda-1)(\lambda+1)^2\).
Expanding \(\det(\lambda I-A)\) along the first row of \[ \lambda I-A=\begin{pmatrix}\lambda&1&-2\\-3&\lambda+4&-6\\-2&2&\lambda-3\end{pmatrix} \] gives \[ \lambda(\lambda^2+\lambda)-(-3\lambda-3)-2(2\lambda+2)=\lambda^3+\lambda^2-\lambda-1=(\lambda-1)(\lambda+1)^2. \]
2. The matrix \(A\) is diagonalizable, with minimal polynomial \(m_A(\lambda)=\lambda^2-1\).
The matrix \[ -I-A=\begin{pmatrix}-1&1&-2\\-3&3&-6\\-2&2&-4\end{pmatrix} \] has rank \(1\), since every row is a multiple of \((1,-1,2)\). So the eigenspace for \(-1\) has dimension \(2\), equal to the algebraic multiplicity from step 1, and the eigenvalue \(1\) has multiplicity \(1\). Hence \(A\) is diagonalizable, and its minimal polynomial is the product of the distinct factors \((\lambda-1)(\lambda+1)\).
3. In part (a), the Jordan canonical form is \(J=\operatorname{diag}(1,-1,-1)\), and the rational canonical form is\[ R=\begin{pmatrix}-1&0&0\\0&0&1\\0&1&0\end{pmatrix}, \]with invariant factors \(\lambda+1\mid\lambda^2-1\).
By step 2, \(A\) is similar to the diagonal matrix of its eigenvalues. The invariant factors \(d_1\mid d_2\) satisfy \(d_2=m_A=\lambda^2-1\) and \(d_1d_2=p_A\), so \(d_1=\lambda+1\). Their companion matrices are \((-1)\) and \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), and \(R\) is their block sum [@DF04].
4. In part (b), there are six abelian groups of order \(360\):\[ \ZZ_{360},\quad \ZZ_3\times\ZZ_{120},\quad \ZZ_2\times\ZZ_{180},\quad \ZZ_6\times\ZZ_{60},\quad \ZZ_2\times\ZZ_2\times\ZZ_{90},\quad \ZZ_2\times\ZZ_6\times\ZZ_{30}. \]
Since \(360=2^3\cdot3^2\cdot5\), an abelian group of order \(360\) is \(P_2\times P_3\times P_5\) with \(\abs{P_2}=8\), \(\abs{P_3}=9\), \(\abs{P_5}=5\), and its isomorphism type is determined by a partition of each exponent [@DF04]. The partitions \(3\), \(2+1\), \(1+1+1\) give \(P_2=\ZZ_8\), \(\ZZ_4\times\ZZ_2\), \(\ZZ_2^3\); the partitions \(2\), \(1+1\) give \(P_3=\ZZ_9\), \(\ZZ_3^2\); and \(P_5=\ZZ_5\). The \(3\cdot2\cdot1=6\) combinations, rewritten in invariant-factor form by the Chinese remainder theorem, are the listed groups.
5. Q.E.D.
Step 3 answers part (a) and step 4 answers part (b).