[proofplan]
We establish both directions of the equivalence separately. Since $m_T \mid \chi_T$ (by the [Minimal Polynomial Divides Characteristic Polynomial](/theorems/3291)), every irreducible factor of $m_T$ is automatically an irreducible factor of $\chi_T$. For the converse, we show that if $p(\lambda)$ is an irreducible factor of $\chi_T$ that does not divide $m_T$, then the coprimality of $p$ and $m_T$ leads to the invertibility of $p(T)$, which contradicts the existence of an eigenvector over the algebraic closure. The eigenvalue reformulation follows because the eigenvalues of $T$ are precisely the roots of $\chi_T$ in $k$ (or its extensions).
[/proofplan]
[step:Show every irreducible factor of $m_T$ divides $\chi_T$]
By the [Minimal Polynomial Divides Characteristic Polynomial](/theorems/3291), $m_T \mid \chi_T$ in $k[\lambda]$. Therefore every irreducible factor of $m_T$ is also an irreducible factor of $\chi_T$.
[/step]
[step:Show every irreducible factor of $\chi_T$ divides $m_T$ via a coprimality argument]
Let $p(\lambda) \in k[\lambda]$ be an irreducible polynomial dividing $\chi_T(\lambda)$. Suppose for contradiction that $p(\lambda) \nmid m_T(\lambda)$. Since $p$ is irreducible and does not divide $m_T$, the polynomials $p$ and $m_T$ are coprime in $k[\lambda]$. Since $k[\lambda]$ is a principal ideal domain, there exist polynomials $a, b \in k[\lambda]$ such that
\begin{align*}
a(\lambda) \, m_T(\lambda) + b(\lambda) \, p(\lambda) = 1.
\end{align*}
Substituting the operator $T$ gives
\begin{align*}
a(T) \, m_T(T) + b(T) \, p(T) = I.
\end{align*}
Since $m_T(T) = 0$, this reduces to $b(T) \, p(T) = I$, so $p(T)$ is invertible.
[guided]
The key tool here is Bezout's identity in the PID $k[\lambda]$. Two polynomials over a field are coprime if and only if they generate the unit ideal, which means there exist $a, b \in k[\lambda]$ with $a m_T + b p = 1$. Since $p$ is irreducible, $\gcd(p, m_T) \in \{1, p\}$. If $p \nmid m_T$, then $\gcd(p, m_T) = 1$, so Bezout gives the identity above.
Evaluating at $T$ is valid because polynomial evaluation is a ring homomorphism from $k[\lambda]$ to $\operatorname{End}(V)$: it preserves addition and multiplication. After evaluation, $m_T(T) = 0$ by definition, so the identity becomes
\begin{align*}
a(T) \cdot 0 + b(T) \, p(T) = I,
\end{align*}
which simplifies to $b(T) \, p(T) = I$. This exhibits $b(T)$ as a two-sided inverse of $p(T)$ (since $p(T) \, b(T) = I$ follows by the commutativity of polynomial evaluation), making $p(T)$ an invertible linear operator on $V$.
[/guided]
[/step]
[step:Produce an eigenvector over $\overline{k}$ annihilated by $p(T)$]
Since $p \mid \chi_T$, let $\lambda_0 \in \overline{k}$ be a root of $p$ in the algebraic closure of $k$. Then $\chi_T(\lambda_0) = 0$, so $\det(\lambda_0 I - T_{\overline{k}}) = 0$, where $T_{\overline{k}}: V \otimes_k \overline{k} \to V \otimes_k \overline{k}$ is the base change of $T$ to $\overline{k}$. Therefore $\lambda_0$ is an eigenvalue of $T_{\overline{k}}$, and there exists a nonzero vector $v \in V \otimes_k \overline{k}$ with $T_{\overline{k}} v = \lambda_0 v$.
Evaluating $p$ at $T_{\overline{k}}$ on this eigenvector gives
\begin{align*}
p(T_{\overline{k}}) v = p(\lambda_0) v = 0,
\end{align*}
since $\lambda_0$ is a root of $p$.
[/step]
[step:Derive a contradiction from the invertibility of $p(T)$]
From the coprimality step, $p(T)$ is invertible with inverse $b(T)$. Base-changing to $\overline{k}$, the identity $b(T) \, p(T) = I$ becomes $b(T_{\overline{k}}) \, p(T_{\overline{k}}) = I$ on $V \otimes_k \overline{k}$, so $p(T_{\overline{k}})$ is also invertible. But we have just shown that $p(T_{\overline{k}}) v = 0$ for a nonzero vector $v \in V \otimes_k \overline{k}$, contradicting the invertibility of $p(T_{\overline{k}})$.
Therefore our assumption $p \nmid m_T$ is false, and $p \mid m_T$.
[guided]
Let us trace through why the base change preserves invertibility.
The Bezout identity $b(T) \, p(T) = I$ is an equation in $\operatorname{End}_k(V)$.
Applying the base-change functor $- \otimes_k \overline{k}$ to both sides (which is a ring homomorphism on endomorphism algebras), we obtain
\begin{align*}
b(T_{\overline{k}}) \, p(T_{\overline{k}}) = I_{\overline{k}}
\end{align*}
in $\operatorname{End}_{\overline{k}}(V \otimes_k \overline{k})$. In concrete terms: if $p(T)$ has an inverse matrix over $k$, the same matrix is still an inverse over the larger field $\overline{k}$, so $p(T_{\overline{k}})$ is invertible with trivial kernel.
On the other hand, the eigenvector $v \neq 0$ satisfies $p(T_{\overline{k}}) v = p(\lambda_0) v = 0$ (because $\lambda_0$ is a root of $p$), so $v \in \ker p(T_{\overline{k}})$.
A nonzero vector in the kernel of an invertible operator is impossible, so we have a contradiction.
Therefore the assumption $p \nmid m_T$ is false, and we conclude $p \mid m_T$.
[/guided]
[/step]
[step:Conclude the eigenvalue characterisation]
Combining the two directions: an irreducible polynomial $p \in k[\lambda]$ divides $\chi_T$ if and only if $p$ divides $m_T$. In other words, $m_T$ and $\chi_T$ have the same irreducible factors in $k[\lambda]$.
For the eigenvalue reformulation: $\lambda_0 \in k$ is an eigenvalue of $T$ if and only if $\chi_T(\lambda_0) = \det(\lambda_0 I - T) = 0$, which holds if and only if $(\lambda - \lambda_0)$ divides $\chi_T(\lambda)$. Since $(\lambda - \lambda_0)$ is irreducible in $k[\lambda]$, this occurs if and only if $(\lambda - \lambda_0)$ divides $m_T(\lambda)$.
[/step]
