[proofplan]
We use the Cayley-Hamilton theorem, which asserts $\chi_T(T) = 0$, together with the defining property of the minimal polynomial. Since $m_T$ is the monic polynomial of least degree annihilating $T$, and $\chi_T$ also annihilates $T$, we perform polynomial division of $\chi_T$ by $m_T$ and show the remainder must vanish. The degree inequality then follows from the divisibility relation and the fact that $\deg \chi_T = \dim V$.
[/proofplan]
[step:Divide $\chi_T$ by $m_T$ in $k[\lambda]$ with remainder]
Since $k[\lambda]$ is a Euclidean domain, we apply the division algorithm to write
\begin{align*}
\chi_T(\lambda) = q(\lambda) \, m_T(\lambda) + r(\lambda),
\end{align*}
where $q, r \in k[\lambda]$ and either $r = 0$ or $\deg r < \deg m_T$.
[/step]
[step:Evaluate at $T$ and apply the Cayley-Hamilton theorem to show $r(T) = 0$]
Substituting the operator $T$ into the division identity gives
\begin{align*}
\chi_T(T) = q(T) \, m_T(T) + r(T).
\end{align*}
By the Cayley-Hamilton theorem, $\chi_T(T) = 0$. By definition of the minimal polynomial, $m_T(T) = 0$. Therefore
\begin{align*}
0 = q(T) \cdot 0 + r(T) = r(T).
\end{align*}
[guided]
The Cayley-Hamilton theorem states that every linear operator on a finite-dimensional vector space is annihilated by its own characteristic polynomial: $\chi_T(T) = 0$. We substitute the operator $T$ into both sides of the polynomial division identity $\chi_T(\lambda) = q(\lambda) \, m_T(\lambda) + r(\lambda)$. This substitution is valid because polynomial evaluation $k[\lambda] \to \operatorname{End}(V)$, $f(\lambda) \mapsto f(T)$, is a ring homomorphism — it preserves addition and multiplication. On the left-hand side, the Cayley-Hamilton theorem gives $\chi_T(T) = 0$. On the right-hand side, $m_T(T) = 0$ by the very definition of the minimal polynomial (it is the monic polynomial of least degree satisfying $m_T(T) = 0$). Therefore
\begin{align*}
0 = \chi_T(T) = q(T) \, m_T(T) + r(T) = q(T) \cdot 0 + r(T) = r(T).
\end{align*}
We conclude that the remainder polynomial $r$ also annihilates $T$: $r(T) = 0$.
[/guided]
[/step]
[step:Conclude $r = 0$ from the minimality of $m_T$]
We have $r(T) = 0$ with either $r = 0$ or $\deg r < \deg m_T$. If $r \neq 0$, then $r$ would be a nonzero polynomial of degree strictly less than $\deg m_T$ that annihilates $T$. After rescaling $r$ to be monic, this would contradict the definition of $m_T$ as the monic polynomial of least degree annihilating $T$. Therefore $r = 0$, and the division identity becomes
\begin{align*}
\chi_T(\lambda) = q(\lambda) \, m_T(\lambda),
\end{align*}
so $m_T \mid \chi_T$ in $k[\lambda]$.
[guided]
Why must $r$ vanish? Recall that the minimal polynomial $m_T$ is defined as the monic polynomial of least degree in $k[\lambda]$ satisfying $m_T(T) = 0$.
We have established that $r(T) = 0$, and the division algorithm guarantees that either $r = 0$ or $\deg r < \deg m_T$.
Suppose for contradiction that $r \neq 0$. Then $r$ is a nonzero polynomial annihilating $T$ with $\deg r < \deg m_T$.
Let $c \in k^\times$ be the leading coefficient of $r$, and define $\tilde{r} := c^{-1} r$. Then $\tilde{r}$ is monic, $\tilde{r}(T) = c^{-1} r(T) = 0$, and $\deg \tilde{r} = \deg r < \deg m_T$.
This contradicts the minimality of $m_T$: we have found a monic annihilating polynomial of strictly smaller degree. Therefore $r = 0$.
With $r = 0$, the division identity $\chi_T = q \cdot m_T + r$ becomes
\begin{align*}
\chi_T(\lambda) = q(\lambda) \, m_T(\lambda),
\end{align*}
which is precisely the statement that $m_T$ divides $\chi_T$ in $k[\lambda]$.
[/guided]
[/step]
[step:Deduce the degree bound $\deg m_T \le \dim V$]
Since $m_T \mid \chi_T$, we have $\deg m_T \le \deg \chi_T$. The characteristic polynomial $\chi_T(\lambda) = \det(\lambda I - T)$ is the determinant of an $n \times n$ matrix (where $n = \dim V$), so $\deg \chi_T = n = \dim V$. Therefore
\begin{align*}
\deg m_T \le \deg \chi_T = \dim V.
\end{align*}
[/step]
