Proof that the eigenvalues are the roots of the minimal polynomial

Proof

We prove the two implications separately.

1. Let $m_T$ be the minimal polynomial of $T$, and write $\id = \id_V$.
2. First let $\lambda$ be an eigenvalue of $T$. So there is $v \in V$ with $v \ne 0$ and $T(v) = \lambda v$.
3. By induction on $k$, $T^k(v) = \lambda^k v$ for all $k \ge 0$: the case $k = 0$ is $\id(v) = v$, and $T^{k+1}(v) = T(\lambda^k v) = \lambda^{k+1} v$.
4. Write $m_T(x) = \sum_{k=0}^{d} c_k x^k$. By the definition of the value at $T$ and step 3, $m_T(T)(v) = \sum_k c_k \lambda^k v = m_T(\lambda) v$.
5. Since $m_T(T) = 0$, step 4 gives $m_T(\lambda) v = 0$. As $v \ne 0$, we get $m_T(\lambda) = 0$.
6. Conversely, let $m_T(\lambda) = 0$. By the factor theorem, $m_T = (x - \lambda) q$ for some $q \in F[x]$.
7. The polynomial $q$ is monic, and its degree is $\deg m_T - 1$. By the minimality of $\deg m_T$, $q(T) \ne 0$.
8. So there is $w \in V$ with $u = q(T)(w) \ne 0$.
9. Evaluation at $T$ respects products, so $(T - \lambda \id)(u) = \big((x - \lambda) q\big)(T)(w) = m_T(T)(w) = 0$.
10. Hence $T(u) = \lambda u$ with $u \ne 0$, and $\lambda$ is an eigenvalue of $T$.
11. Steps 5 and 10 show that $\lambda$ is an eigenvalue of $T$ if and only if $m_T(\lambda) = 0$.