Rational canonical form of diagonal matrix

Question

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.

Answer 1

Yes, the rational canonical form is just $A$. Your intuition is good, but can be expounded futher:

Let $V$ be a vector space where the matrix of some linear operator $T$ is represented by $A$ in some basis. Since $A$ is diagonal, $T$ is a diagonalizable operator, so $V$ has a basis where each vector is an eigenvector for $T$. The Cyclic Decomposition Theorem lets us decompose $V$ into a direct sum of $T$-cyclic subspaces for some vectors $\alpha_1,...,\alpha_k$. Namely, we can decompose $V$ as:

$$V = Z(\alpha_1; T) \oplus ... \oplus Z(\alpha_n; T)$$

Where $Z(\alpha_i; T) = \{ v \in V: v = g(T) \alpha_i \ \text{ for some polynomial } \ g(x) \}$, and $\alpha_1,...,\alpha_n$ are the distinct eigenvectors of $A$.

These subspaces are invariant under $T$, so let $T_i$ be the operator induced by $T$ on $Z(\alpha_i; T)$. Then $T_i$ has minimal polynomial $p_i(x) = x - a_i$, where $a_i$ is the eigenvalue corresponding to the eigenvector $\alpha_i$. Since $\dim(Z(\alpha_i; T)) = 1$ and $T_i$ has a cyclic vector, we get that the rational form of $T_i$ is just its $1 \times 1$ companion matrix:

$A_i = \begin{bmatrix} a_i \end{bmatrix}$.

Thus $$A = A_1 \oplus ... \oplus A_n = \begin{bmatrix} a_1 & 0 &... & 0 \\ 0 & a_2 & ... & 0 \\ . \\ . \\ 0 & 0 & ... & a_n \end{bmatrix}$$

Is the rational form of $A$

Answer 2

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]);"