[proofplan] We establish four inequalities. $g_\lambda \geq 1$ from the definition of eigenvalue. $g_\lambda \leq a_\lambda$ by choosing a basis starting with eigenvectors and applying the [Block Triangular Determinant](/theorems/399). $c_\lambda \geq 1$ because evaluating $M_\alpha$ at the eigenvalue via an eigenvector forces $\lambda$ to be a root. $c_\lambda \leq a_\lambda$ because $M_\alpha \mid \chi_\alpha$ by [Cayley-Hamilton](/theorems/407). [/proofplan] [step:Show $g_\lambda \geq 1$ from the definition of eigenvalue] Since $\lambda$ is an eigenvalue, $E_\alpha(\lambda) = \ker(\alpha - \lambda\,\mathrm{id}) \neq \{0\}$ by definition. Hence $g_\lambda = \dim E_\alpha(\lambda) \geq 1$. [/step] [step:Show $g_\lambda \leq a_\lambda$ by block triangular decomposition of $\chi_\alpha$] Let $g = g_\lambda$ and choose a basis $(v_1, \dots, v_g)$ for $E_\alpha(\lambda)$. Extend to a basis $(v_1, \dots, v_g, v_{g+1}, \dots, v_n)$ for $V$. In this basis, the matrix of $\alpha$ has block form \begin{align*} A = \begin{pmatrix} \lambda I_g & C \\ \mathbf{0} & D \end{pmatrix}, \end{align*} since $\alpha(v_i) = \lambda v_i$ for $i = 1, \dots, g$. By the [Block Triangular Determinant](/theorems/399): \begin{align*} \chi_\alpha(t) = \det(tI - A) = (t - \lambda)^g \det(tI_{n-g} - D). \end{align*} The factor $\det(tI_{n-g} - D)$ may contribute additional powers of $(t - \lambda)$. Therefore $a_\lambda \geq g = g_\lambda$. [/step] [step:Show $c_\lambda \geq 1$ by evaluating $M_\alpha$ at an eigenvector] Since $\lambda$ is an eigenvalue, there exists $v \neq \mathbf{0}$ with $\alpha(v) = \lambda v$. For any polynomial $f \in \mathbb{F}[t]$, evaluating at $\alpha$ and applying to $v$ gives $f(\alpha)(v) = f(\lambda)\,v$. In particular, $M_\alpha(\alpha)(v) = M_\alpha(\lambda)\,v$. Since $M_\alpha(\alpha) = 0$ and $v \neq \mathbf{0}$, we conclude $M_\alpha(\lambda) = 0$. Therefore $(t - \lambda) \mid M_\alpha(t)$, giving $c_\lambda \geq 1$. [guided] The identity $f(\alpha)(v) = f(\lambda)\,v$ holds because $\alpha(v) = \lambdav$ implies $\alpha^k(v) = \lambda^k v$ by induction. Then for $f(t) = \sum a_i t^i$: $f(\alpha)(v) = \sum a_i \alpha^i(v) = \sum a_i \lambda^i v = f(\lambda)\,v$. This shows that eigenvectors "see" polynomial evaluations at the eigenvalue. Since $M_\alpha(\alpha) = 0$ annihilates everything, applying it to $v$ gives $M_\alpha(\lambda)\,v = \mathbf{0}$, and $v \neq \mathbf{0}$ forces $M_\alpha(\lambda) = 0$. [/guided] [/step] [step:Show $c_\lambda \leq a_\lambda$ from $M_\alpha \mid \chi_\alpha$] By [Cayley-Hamilton](/theorems/407), $M_\alpha(t) \mid \chi_\alpha(t)$. If $(t - \lambda)^{c_\lambda} \mid M_\alpha(t)$ and $M_\alpha(t) \mid \chi_\alpha(t)$, then $(t - \lambda)^{c_\lambda} \mid \chi_\alpha(t)$. Since $a_\lambda$ is the exact power of $(t - \lambda)$ dividing $\chi_\alpha(t)$, we conclude $c_\lambda \leq a_\lambda$. [/step]