I want to take this question as an example:
Rational canonical form(Frobenius normal form) of the matrix $A$
The matrix looks like:
$$ A=\left[\begin{array}{ccc} 2 & 1 & 2 \\ -2 & -1 & -4 \\ 1 & 1 & 3 \end{array}\right] $$
The rational canonical form (Frobenius normal form) is:
$$ R=\left[\begin{array}{ccc} 0 & -2 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] $$
It is said that the two will be similar. That is, $A=S^{-1}RS$ for some non-singular $S$. My questions are (the first one is kind of silly but it really confused me):
- The other way around. If I can find some matrix $S$ such that $A=S^{-1}RS$, can I guarantee that $S$ is invertible?
- (Main question) How do I find a non-singular matrix $S$ to make the equation $A=S^{-1}RS$ hold.
Currently, I assume $S$ takes the form
$$ S=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] $$
Then do $A=S^{-1}RS\Rightarrow SA=RS$, that is,
$$ \left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \left[\begin{array}{ccc} 2 & 1 & 2 \\ -2 & -1 & -4 \\ 1 & 1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 0 & -2 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] $$
But I cannot solve $S$ out......
Then, the question 2 becomes how to find the $ S=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] $.
And also, I think there might be some method using the idea of decompositions can solve this much faster, I just cannot figure it out.
It would be helpful if anyone has some hint on the problem. Thank you all very much.