[proofplan]
We prove the LCM identity by showing divisibility in both directions. First, since $m_T(T) = 0$ annihilates every vector, $m_{T,v} \mid m_T$ for all $v$, so $\operatorname{lcm}_v m_{T,v} \mid m_T$. Conversely, the LCM annihilates every vector and hence annihilates $T$, so $m_T \mid \operatorname{lcm}_v m_{T,v}$. For the existence of a vector achieving $m_{T,v} = m_T$, we use the rational canonical form: a cyclic direct summand of maximal invariant-factor degree provides such a vector.
[/proofplan]
[step:Show $m_{T,v} \mid m_T$ for every nonzero $v \in V$]
Let $v \in V \setminus \{0\}$. The minimal polynomial $m_{T,v} \in k[\lambda]$ is defined as the monic polynomial of least degree satisfying $m_{T,v}(T) v = 0$. Since $m_T(T) = 0$ (the zero operator on $V$), we have $m_T(T) v = 0$ in particular. Performing polynomial division, write $m_T = q \cdot m_{T,v} + r$ with $\deg r < \deg m_{T,v}$ or $r = 0$. Then $r(T) v = m_T(T) v - q(T) m_{T,v}(T) v = 0 - 0 = 0$. By minimality of $\deg m_{T,v}$, we must have $r = 0$, so $m_{T,v} \mid m_T$.
Since this holds for every nonzero $v$, the least common multiple satisfies
\begin{align*}
\operatorname{lcm}_{v \in V \setminus \{0\}} m_{T,v} \mid m_T.
\end{align*}
[guided]
The vector minimal polynomial $m_{T,v}$ captures the annihilating behaviour of $T$ on a single vector: it is the monic generator of the ideal $\{f \in k[\lambda] : f(T) v = 0\}$ in the PID $k[\lambda]$. Since $m_T(T) = 0$ as an operator on $V$, the polynomial $m_T$ annihilates every vector, so $m_T$ belongs to this ideal for each $v$. Because $m_{T,v}$ generates the ideal, we have $m_{T,v} \mid m_T$.
More explicitly, the division algorithm in $k[\lambda]$ gives $m_T = q \cdot m_{T,v} + r$ with $\deg r < \deg m_{T,v}$ or $r = 0$. Evaluating at $T$ and applying to $v$:
\begin{align*}
r(T) v = m_T(T) v - q(T) \, m_{T,v}(T) v = 0 - 0 = 0.
\end{align*}
If $r \neq 0$, then after making $r$ monic we obtain a polynomial of degree $< \deg m_{T,v}$ that annihilates $v$ under $T$, contradicting the minimality of $m_{T,v}$. So $r = 0$ and $m_{T,v} \mid m_T$.
Taking the LCM over all nonzero vectors preserves the divisibility: since each $m_{T,v}$ divides $m_T$, their LCM also divides $m_T$.
[/guided]
[/step]
[step:Show $m_T \mid \operatorname{lcm}_v m_{T,v}$ by proving the LCM annihilates $T$]
Let $L := \operatorname{lcm}_{v \in V \setminus \{0\}} m_{T,v}$. For every nonzero $v \in V$, we have $m_{T,v} \mid L$, so $L(T) v = 0$. For $v = 0$, $L(T) v = L(T) \cdot 0 = 0$ as well. Therefore $L(T) v = 0$ for all $v \in V$, which means $L(T) = 0$ as a linear operator on $V$.
Since $m_T$ is the monic polynomial of least degree annihilating $T$, and $L$ annihilates $T$, we have $m_T \mid L$ by the same division-algorithm argument as in the previous step.
[/step]
[step:Conclude the LCM identity]
From the previous two steps, $\operatorname{lcm}_v m_{T,v} \mid m_T$ and $m_T \mid \operatorname{lcm}_v m_{T,v}$. Since both are monic polynomials, they are equal:
\begin{align*}
m_T = \operatorname{lcm}_{v \in V \setminus \{0\}} m_{T,v}.
\end{align*}
[/step]
[step:Produce a vector $v$ with $m_{T,v} = m_T$ using the rational canonical form]
By the rational canonical form theorem, there exists a decomposition
\begin{align*}
V = C_{f_1} \oplus C_{f_2} \oplus \cdots \oplus C_{f_s},
\end{align*}
where each $C_{f_i}$ is a $T$-cyclic subspace (the $T$-cyclic subspace generated by some vector $w_i$, with $m_{T, w_i} = f_i$), and the invariant factors satisfy $f_1 \mid f_2 \mid \cdots \mid f_s$. The minimal polynomial of $T$ equals the largest invariant factor: $m_T = f_s$.
Let $v := w_s$, the cyclic generator of the last summand. Then $m_{T,v} = f_s = m_T$.
[guided]
Why does the rational canonical form guarantee the existence of such a vector?
The rational canonical form decomposes $V$ into $T$-cyclic subspaces $C_{f_i}$, where $C_{f_i} = \operatorname{span}\{w_i, T w_i, T^2 w_i, \ldots, T^{\deg f_i - 1} w_i\}$ and $f_i(T) w_i = 0$ with $f_i = m_{T, w_i}$.
The invariant factors $f_1 \mid f_2 \mid \cdots \mid f_s$ are uniquely determined by $T$, and the minimal polynomial of $T$ is the largest invariant factor $f_s$.
To verify that $m_T = f_s$, we check both divisibilities. First, $m_T$ must annihilate every cyclic summand $C_{f_i}$, so $f_i \mid m_T$ for all $i$; in particular $f_s \mid m_T$.
Conversely, since the invariant factors satisfy $f_i \mid f_s$ for all $i$, write $f_s = g_i \cdot f_i$. Then $f_s(T) w_i = g_i(T) f_i(T) w_i = 0$ for every generator $w_i$.
Since the $w_i$ generate all of $V$ under powers of $T$, we get $f_s(T) = 0$ on $V$, giving $m_T \mid f_s$.
Combining both divisibilities with the fact that $m_T$ and $f_s$ are both monic yields $m_T = f_s$. The cyclic generator $w_s$ of the last summand then satisfies $m_{T, w_s} = f_s = m_T$, as required.
[/guided]
[/step]
