Show that the rational form of $A$ remains the same whether viewed as a matrix over $\mathbb{C}$ or over $\mathbb{F}$(subfield)

Solution 1:

This question can be quickly answered with an approach framed by the following result.

HK Section 7.2, Theorem 5: Let $\Bbb F$ be a field and let $B$ be an $n \times n$ matrix over $\Bbb F$. Then $B$ is similar over the field $\Bbb F$ to one and only one matrix which is in rational form.

Now, let $A$ be an arbitrary $n \times n$ matrix with elements in $\Bbb F$ (i.e. $A \in \Bbb F^{n \times n}$). We know from the theorem that there exist matrices $C,P \in \Bbb F^{n \times n}$ such that $C$ is in rational canonical form, $P$ is invertible, and $$ A = P^{-1}CP. $$ It follows that when $A,C,P$ are considered as elements of $\Bbb C^{n \times n}$, we still have $A = P^{-1}CP$. However, $C$ is in rational canonical form. By theorem 5, this means that $C$ is the rational canonical form of $A$, as a matrix in $\Bbb C^{n \times n}$.

The conclusion follows.

We are given that $\Bbb F$ is a subfield of $\Bbb C$. For any vector space $V$ over $\Bbb F$, I will use $\bar V$ to refer to the extension of $V$ to a vector space over $\Bbb C$. In your setting, we have $V \subset \Bbb F^n \subset \Bbb C^n$, so we can more simply say that $\bar V = \{z v : z \in \Bbb C, v \in V\}$. In a more general setting, we might say that $\bar V = V \otimes_{\Bbb F} \Bbb C$ is obtained via an extension of scalars.

Let $T$ denote the linear map $T:\Bbb F^n \to \Bbb F^n$ given by $T(x) = Ax$. In a similar vein, I will use $\bar T$ to refer to the map induced by $T$ on $\Bbb C^n$, whereas $T$ refers to a map on $\Bbb F^n$. More generally: for a $\Bbb F$-linear map $\tau:V \to V$, $\bar \tau$ denotes the induced $\Bbb C$-linear map $\bar \tau : \bar V \to \bar V$

I claim (without proof) that the the following holds:

1. If $V = V_1 \oplus \cdots \oplus V_k$, then $\bar V = \bar V_1 \oplus \cdots \oplus \bar V_k$
2. $\overline{Z_{\Bbb F}(\alpha,T)} = Z_{\Bbb C}(\alpha,\bar T)$.

The subscripts here emphasize $Z(\alpha,T) = Z_{\Bbb F}(\alpha,T)$ is a vector space over $\Bbb F$, whereas $Z(\alpha,\bar T) = Z_{\Bbb C}(\alpha,\bar T)$ is a vector space over $\Bbb C$.

We are given that $\Bbb F^n$ has the $T$-cyclic decomposition $$ \Bbb F^n = Z(\alpha_1;T) \oplus \cdots \oplus Z(\alpha_k;T) $$ where $p_{i+1}\mid p_i$ for $i = 1,\dots,k-1$. The question asks us to show that $\bar T$ has the same canonical form as $T$. To do this, it suffices to show that $\Bbb C^n$ has the $\bar T$-cyclic decomposition $$ \Bbb C^n = Z(\alpha_1;\bar T)\oplus \cdots \oplus Z(\alpha_k;\bar T) $$ and that the $\bar T$-annihilator of $\alpha_i$ is $p_i$ for each $i$.

First of all, using statements 1 and 2, we have $$ \Bbb C^n = \overline{\Bbb F^n} = \overline{Z(\alpha_1;T) \oplus \cdots \oplus Z(\alpha_k;T)} = Z(\alpha_1;\bar T)\oplus \cdots \oplus Z(\alpha_k;\bar T). $$ Now, it suffices to show that for each $i$, the $\bar T$-annihilator of $\alpha_i$ is equal to $p_i$, the $T$-annihilator of $\alpha_i$. To show that this holds, note that the $T$-annihilator of $\alpha_i$ is the characteristic/minimal polynomial of $T|_{Z(\alpha_i;T)}$. From your previous question, we know that $\overline{T|_{Z(\alpha_i; T)}}$ has the same characteristic and minimal polynomials as $T|_{Z(\alpha_i; T)}$. However, $$ \overline{T|_{Z(\alpha_i; T)}} = \bar T|_{\overline{Z(\alpha_i;T)}} = \bar T|_{Z(\alpha_i;\bar T)}, $$ which means that the $\bar T$-annihilator of $\alpha_i$ is equal to the minimal polynomial of $\bar T|_{Z(\alpha_i;\bar T)}$, which is equal to the minimal polynomial of $T|_{Z(\alpha_i; T)}$, which is equal to $p_i$.

The conclusion follows.