Irreducible factors of minimal and characteristic polynomial of a endomorphism over a finite dimensional $\mathbb{F}$-vector space

Question

Let $V$ be a finite dimensional $\mathbb{F}$-vector space. Suppose $L:V\to V$ is an endomorphism, whose associated matrix is $A$. Now, denote its characteristic and minimal polynomial by \begin{align*} p_L(\lambda) = \det(A - \lambda \mathbb{I}) = \prod_{i=1}^r p_i(\lambda)^{e_i} \\ m_l(\lambda) = \prod_{i=1}^r p_i(\lambda)^{m_i} \end{align*} where $p_i(\lambda) \in \mathbb{F}[\lambda]$ are monic and irreducible over $\mathbb{F}$ and $e_i \in \mathbb{N}, e_i \geq m_i \in \mathbb{N} \cup \{0\}$. We know we can do this by Cayley-Hamilton theorem. Is there an elementary way to prove that $m_i \in \mathbb{N}$ without using any Field Theory?. I know, by Bezout's Lemma, that \begin{align*} V = \bigoplus_{i=1}^r \ker[p_i(L)^{e_i}] = \bigoplus_{i=1}^r \ker[p_i(L)^{m_i}] \end{align*} The original statement is equivalent to the following statements \begin{gather*} \det(p_i(A)) = 0 \\ \det(p_i(A)^{e_i}) = 0 \\ \dim(\ker[p_i(L)^{e_i}]) = \deg(p_i(\lambda))^{e_i} \\ \dim(\ker[p_i(L)^{e_i}]) \leq \deg(p_i(\lambda))^{e_i} \\ \dim(\ker[p_i(L)^{e_i}]) \geq \deg(p_i(\lambda))^{e_i} \\ p_{L_{|\ker[p_i(L)^{e_i}]}} = p_i(L)^{e_i} \\ p_{L_{|\ker[p_i(L)^{e_i}]}} = p_i(L)^{a} \end{gather*}

Answer 1

The "tricky part", I think, is showing that $p_i(L)$ must have a non-trivial kernel.

Here's an approach that uses field theory, but without leaning on the usual statement that "there exists an algebraic closure, therefore we can pretend that the field was algebraically closed to begin with".

Suppose that $p(\lambda)$ is a non-linear, irreducible factor of the characteristic polynomial of $L$. Consider the field $\Bbb K = \Bbb F[x]/\langle p(x)\rangle$, and let $\tilde L$ denote the map induced by $L$ on the $\Bbb K$-space $V \otimes_{\Bbb F} \Bbb K$. That is, $\tilde L$ is the unique $\Bbb K$-linear map for which $$ \tilde L(v \otimes_{\Bbb F} k) = L(v) \otimes_{\Bbb F} k, \quad \text{for all } v \in V, k \in \Bbb K. $$ We find that $\det(\tilde L - \lambda I) = [\det(L - \lambda I)] \otimes_{\Bbb F} 1$. It follows that $\det(\tilde L - \lambda I)$, as a polynomial over $\Bbb K$, has $x \in \Bbb K$ as a root. That is, $\det(\tilde L - x I) = 0$. Consequently, there exists a non-zero vector $v_x \in V \otimes_{\Bbb F} \Bbb K$ such that $ (\tilde L - x I)v_x = 0 $. Because $p(x) = 0$, we can conclude from this that $p(\tilde L)v_x = 0$.

Now, suppose that $p(\lambda) = a_0 + a_1 \lambda + \cdots + a_{d-1} \lambda^{d-1} + \lambda^d$, so that all elements of $\Bbb K$ can be expressed in the form $b_0 + b_1 x + \cdots + b_{d-1}x^{d-1}$ for $b_0,\dots,b_{d-1} \in \Bbb F$. Write $v_x$ as $$ v_x = v_0 \otimes 1 + v_1 \otimes x + \cdots + v_{d-1} \otimes x^{d-1}, \quad v_0,\dots,v_{d-1} \in V. $$ The fact that $p(\tilde L)v_x = 0$ can be expanded to yield $$ p(L)v_0 \otimes 1 + p(L)v_1 \otimes x + \cdots + p(L)v_{d-1} \otimes x^{d-1} =0. $$ In other words, we have $p(L)v_i = 0$ for $i = 0,1,\dots,d-1$. Because $v_x$ was a non-zero vector, at least one of the vectors $v_i \in V$ must be non-zero. Thus, we conclude that the kernel of $p(L)$ must be non-trivial.

---

From there, suppose for the purpose of contradiction that $p(\lambda)$ does not divide the minimal polynomial $m_L(\lambda)$ of $L$. By Bezout's lemma, there exist polynomials $f(\lambda),g(\lambda)$ such that $$ f(\lambda)p(\lambda) + g(\lambda)m_L(\lambda) = 1. $$ Correspondingly, we have $$ f(L)p(L) + g(L)m_L(L) = I. $$ However, we must have $m_L(L) = 0$. So, we have $f(L)p(L) = I$, which is to say that $f(L)$ is the inverse of $p(L)$. This is impossible: $p(L)$ has a non-trivial kernel, and therefore does not have an inverse.

Conclude that $p$ divides $m_L$.

Answer 2

The claim is automatically true for any Companion Matrix, $C$, since the minimal polynomial $m$ of $C$ is equal to the characteristic polynomial $p$ of $C$. The fact that $m=p$ is elementary since it is provable by a simple calculation. (Technically I require that the characteristic polynomial is monic.)

The below proves the equivalent claim that $\det(p_i(A)) = 0$. This is really a claim about polynomials so rudimentary ring theory seems appropriate. However no knowledge of Field Theory, e.g. Field Extensions is required.

---

The argument is essentially that a polynomial $q$ applied to a matrix $A$ and its Companion Matrix $C$ gives matrix $q\big(A\big)$ and $q\big(C\big)$ with the same characteristic polynomial and hence same determinant. Setting $q:= p_i$ then allows us to relate $\det\big(p_i(A)\big)=\det\big(p_i(C)\big)=0$ by the opening paragraph.

To prove $q\big(A\big)$ and $q\big(C\big)$ have the same characteristic polynomial when working over any field it is convenient to work in a slightly more general setting of $n\times n$ matrices with components from the ring ring $R:=\mathbb Z\big[\mathbf x\big]$ (which is an integral domain of characteristic zero so Newton's Identities are maximally effective here).
$\mathbf x := \left[\begin{matrix}x_1 \\ x_2\\ \vdots \\x_{n^2}\end{matrix}\right]$

and consider $A \in R^{n\times n}$
where $a_{1,1}= x_1$, $a_{2,1}= x_2$,... , $a_{n,1}= x_n$, $a_{1,2}=x_{n+1}$ and so on. And let $C \in R^{n\times n}$ be its Companion Matrix.

Then $A$ and $C$ have the same characteristic polynomial, so
$\text{trace}\big(A^k\big)=\text{trace}\big(C^k\big)$ for $k\in \big\{1,2,...\big\}$ by Newton's Identities

By Cayley-Hamilton, any polynomial $q_\alpha$ of $C$ (or $A$) may be written as
$q_\alpha\big(C\big) = \gamma_{\alpha_0}I_n+\sum_{r=1}^{n}{\alpha_r}C^r$
(which is in fact unique for the Companion Matrix though we don't really need this) and
$q_\alpha\big(A\big) = \gamma_{\alpha_0}I_n+\sum_{r=1}^{n}{\alpha_r}A^r$
(in general this is not unique) Thus
$\text{trace}\Big(q_\alpha\big(C\big)\Big) = \gamma_{\alpha_0}\cdot n+\sum_{r=1}^{n}{\alpha_r}\cdot\text{trace}\Big(C^r\Big)=\gamma_{\alpha_0}\cdot n+\sum_{r=1}^{n}{\alpha_r}\cdot\text{trace}\Big(A^r\Big)=\text{trace}\Big(q_\alpha\big(A\big)\Big)$

$\implies\text{trace}\Big(\big(q_\alpha(C)\big)^k\Big)=\text{trace}\Big(\big(q_\alpha(A)\big)^k\Big)$
because $\big(q_\alpha(C)\big)^k=q_{\alpha'}(C)$ i.e. a power of a polynomial of $C$ (or $A$) is another polynomial of $C$ (or $A$).

By Newton's Identities $q_\alpha\big(C\big)$ and $q_\alpha\big(A\big)$ have the same characteristic polynomial, and in particular
$\det\Big(q_\alpha\big(C\big)\Big)=\det\Big(q_\alpha\big(A\big)\Big)$

Now consider the ring substitution homomorphism
$\phi:R \longrightarrow \mathbb F$
which sends $\mathbb Z$ to the ground field $\mathbb F_p$ (or $\mathbb Q$ if characteristic zero) and $x_1 \mapsto$ the desired value for $a_{1,1}\in \mathbb F$ and $x_2\mapsto$ the desired value for $a_{2,1} \in \mathbb F$ and so on.

Then define
$\Phi: R^{n\times n} \longrightarrow \mathbb F^{n\times n}$
which applies $\phi$ component-wise.

Since the determinant is a polynomial in the entries of a matrix, for arbitrary $B \in R^{n\times n}$
$\det\Big(\Phi\big(B\big)\Big)= \phi\Big(\det\big(B\big)\Big)$ and
$q\big(\Phi(B)\big)= \Phi\big(q(B)\big)$
since each component of $q(B)$ is a polynomial in the entries of $B$ thus $\phi\big(q(B)_{i,j}\big)=q\big(\phi(B)_{i,j}\big)$

and selecting $B:=q(A)$
$\det\Big(q\big(\Phi(A)\big)\Big)= \det\Big(\Phi\big(q(A)\big)\Big)= \phi\Big(\det\big(q(A)\big)\Big) $
$=\phi\Big(\det\big(q(C)\big)\Big)=\det\Big(\Phi\big(q(C)\big)\Big)=\det\Big(q\big(\Phi(C)\big)\Big)$
which completes the proof