I'm trying to determine the rational canonical form of a diagonal matrix
$$
A=\begin{pmatrix}
a_1 & 0 & \cdots & 0\\
0 & a_2 & \cdots & 0\\
\vdots & \vdots & & \vdots\\
0 &0 & \cdots & a_n
\end{pmatrix}
$$ where the $a_i$'s are all different. If my intuition is correct, since the characteristic polynomial (in this case also the minimal polynomial) of $A$ is just the product $(x-a_1)\cdots(x-a_n)$ and all the factors are different we have that the $(x-a_i)$'s are the invariant factors of $A$. Then the rational canonical form of $A$ is again $A$.
Is this correct? Is there a more formal way to work this problem? I'd appreciate any suggestions. Thanks in advance.
Best Answer
I totally disagree with the answers given previously. The authors confuse the Frobenius normal form with the primary rational canonical form.
The Frobenius decomposition has the following form
$F:=diag(C_{p_1},\cdots,C_{p_k})$ where the $C_{p_i}$ are the companion matrices of the polynomial $p_i$, and overall, $p_i$ is a divisor of $p_{i+1}$. In particular, $p_k$ is the minimal polynomial and $p_1\cdots p_k$ is the characteristic polynomial of $A$.
When the eigenvalues of $A$ are distinct, then the vector $(1,\cdots,1)$ is cyclic over whole vector space $K^n$.
Then the Frobenius form of $A$ is $F=C_p$ where $p$ is the characteristic polynomial of $A$.
It suffices to convince oneself to test in Maple
"FrobeniusForm (DiagonalMatrix ([1,2,3]);"