[proofplan] We use the pairwise coprimality of the factors $(t - \lambda_i)^{c_i}$ in $M_\alpha(t)$ to construct projection operators via the Chinese Remainder Theorem (iterated Bezout identities). Each projection maps into a generalised eigenspace $V_i = \ker((\alpha - \lambda_i\,\mathrm{id})^{c_i})$, and the Bezout identity $\sum \pi_i = \mathrm{id}$ yields both the spanning property and directness. [/proofplan] [step:Set up the coprime factorisation and Bezout identity] Write $M_\alpha(t) = \prod_{i=1}^k p_i(t)$ where $p_i(t) = (t - \lambda_i)^{c_i}$ and $\lambda_1, \dots, \lambda_k$ are distinct. Set $V_i = \ker(p_i(\alpha)) = \ker((\alpha - \lambda_i\,\mathrm{id})^{c_i})$. For each $i$, define $q_i(t) = M_\alpha(t)/p_i(t) = \prod_{j \neq i} p_j(t)$. Since the $\lambda_i$ are distinct, $p_i$ and $q_i$ share no roots, hence $\gcd(p_i, q_i) = 1$. By iterating pairwise Bezout identities, there exist $f_1, \dots, f_k \in \mathbb{F}[t]$ with \begin{align*} \sum_{i=1}^k f_i(t)\,q_i(t) = 1. \end{align*} [guided] Why does the simultaneous Bezout identity hold? For $k = 2$, this is standard: $\gcd(p_1, p_2) = 1$ gives $a_1 p_1 + a_2 p_2 = 1$, which is $f_1 q_1 + f_2 q_2 = 1$ since $q_1 = p_2$ and $q_2 = p_1$. For general $k$, use induction: $\gcd(p_1, p_2 \cdots p_k) = 1$ (since $p_1$ is coprime to each $p_j$), so there exist polynomials with $f_1 q_1 + g \cdot p_1 = 1$. Then $g \cdot p_1 = g \cdot (M_\alpha/q_1)$, and we expand $g$ further using the coprimality of $p_2, \dots, p_k$. [/guided] [/step] [step:Construct projection operators and show $\mathrm{im}(\pi_i) \subseteq V_i$] Define $\pi_i = f_i(\alpha) \circ q_i(\alpha) \in \mathrm{End}(V)$. The Bezout identity evaluated at $\alpha$ gives $\sum_{i=1}^k \pi_i = \mathrm{id}$. For any $v \in V$, we have $p_i(\alpha)(\pi_i(v)) = p_i(\alpha)\,f_i(\alpha)\,q_i(\alpha)(v) = f_i(\alpha)\,M_\alpha(\alpha)(v) = \mathbf{0}$, since $p_i \cdot q_i = M_\alpha$ and all polynomial expressions in $\alpha$ commute. Therefore $\pi_i(v) \in \ker(p_i(\alpha)) = V_i$. [/step] [step:Show $V = V_1 + \cdots + V_k$] For any $v \in V$: \begin{align*} v = \mathrm{id}(v) = \sum_{i=1}^k \pi_i(v), \end{align*} and each $\pi_i(v) \in V_i$. So $V = V_1 + \cdots + V_k$. [/step] [step:Show the sum is direct by verifying $\pi_j$ kills $V_i$ for $i \neq j$] For $i \neq j$ and $v_i \in V_i$: $p_i(\alpha)(v_i) = \mathbf{0}$ by definition. Since $i \neq j$, the polynomial $p_i$ divides $q_j = \prod_{\ell \neq j} p_\ell$. Therefore $q_j(\alpha)(v_i) = (\text{polynomial in } \alpha) \circ p_i(\alpha)(v_i) = \mathbf{0}$, so $\pi_j(v_i) = f_j(\alpha)\,q_j(\alpha)(v_i) = \mathbf{0}$. Now suppose $v_1 + \cdots + v_k = \mathbf{0}$ with $v_i \in V_i$. Applying $\pi_j$: $\sum_{i=1}^k \pi_j(v_i) = \pi_j(\mathbf{0}) = \mathbf{0}$. Since $\pi_j(v_i) = \mathbf{0}$ for $i \neq j$, this reduces to $\pi_j(v_j) = \mathbf{0}$. But from $\sum_i \pi_i = \mathrm{id}$ applied to $v_j \in V_j$: $v_j = \sum_i \pi_i(v_j) = \pi_j(v_j)$ (since all other terms vanish). So $v_j = \pi_j(v_j) = \mathbf{0}$ for each $j$. The sum is direct. [guided] The key insight is that the projections $\pi_i$ are mutually orthogonal idempotents: $\pi_i \circ \pi_j = 0$ for $i \neq j$, and $\sum \pi_i = \mathrm{id}$. This follows from $\pi_j(v_i) = 0$ for $v_i \in V_i$ with $i \neq j$, and $\pi_j|_{V_j} = \mathrm{id}_{V_j}$. The latter holds because applying $\sum_i \pi_i = \mathrm{id}$ to $v_j \in V_j$ and using $\pi_i(v_j) = 0$ for $i \neq j$ gives $\pi_j(v_j) = v_j$. These projections decompose every vector uniquely as $v = \sum \pi_i(v)$, and this decomposition is the generalised eigenspace decomposition. [/guided] [/step] [step:Verify the dimension formula $\dim V_i = a_{\lambda_i}$] Since $V = \bigoplus_{i=1}^k V_i$, we have $\sum_{i=1}^k \dim V_i = n$. By [Cayley-Hamilton](/theorems/407), $M_\alpha \mid \chi_\alpha$, so $c_i \leq a_{\lambda_i}$ and each $V_i \supseteq E_\alpha(\lambda_i)$. The restriction $\alpha|_{V_i}$ has only eigenvalue $\lambda_i$ (since $(\alpha - \lambda_j\,\mathrm{id})^{c_j}$ is invertible on $V_i$ for $j \neq i$), so $\chi_{\alpha|_{V_i}}(t) = (t - \lambda_i)^{\dim V_i}$. Since $\chi_\alpha = \prod_i \chi_{\alpha|_{V_i}}$, we get $(t - \lambda_i)^{\dim V_i}$ contributes exactly $\dim V_i$ to the algebraic multiplicity. Therefore $\dim V_i = a_{\lambda_i}$. [/step]