Say we are given the matrix
$$A = \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \\ \end{pmatrix} = \begin{pmatrix} A_{1,1} & A_{1,2} \\ 0 & A_{2,2} \end{pmatrix}$$
and asked to find the Rational Canonical Form. Finding the characteristic polynomial is rather easy considering this is a block matrix. So we have $c_A(x) = \det(xI -A_{1,1})\det(xI-A_{2,2})=(x-2)^6(x-3)^2$.
So my questions are this:
- Is there an easy way of computing the minimal polynomial for this matrix? I know it is $m_A(x) = (x-2)^4(x-3)^2$. However, explicitly computing all combinations of $(A-2I)^n(A-3I)^m$ for $1\leq n \leq 6$ and $1 \leq m \leq 2$ and seeing which smallest one gives us $0$ with a matrix this size (even in block form) seems tedious.
- Given that minimal polynomial, how do we see that the invariant factors are $(x-2), (x-2),$ and $(x-2)^4(x-3)^2$?
Once this is done, the rational canonical form falls out immediately, but I'm unsure of how to show these two questions.
Thanks in advance.
Let $$ R = \left( \begin{array}{cccccccc} 1 & 0& 0 &0& 0& 0& 0& 0 \\ 0 & 0& 0& 0& 1& 0& 0& 0 \\ 0 & 0 &0 &0& 0& 1& 0& 0 \\ 0& 0 &0 &0& 0& 0 &1& 0 \\ 0 &1& 0 &0 &0 &0 &0& 0 \\ 0 &0& 1 &0 &0 &0& 0& 0 \\ 0 &0& 0 &1 &0 &0& 0& 0 \\ 0 &0& 0 &0 & 0 &0 &0 &1 \\ \end{array} \right) $$ then $R^2 = I$ and $RAR$ is in Jordan form