[proofplan] ($\Rightarrow$) If $\alpha$ is diagonalisable with distinct eigenvalues $\lambda_1, \dots, \lambda_k$, the product $p(t) = \prod(t - \lambda_i)$ annihilates $\alpha$ since each factor kills the corresponding eigenspace. By [Existence and Uniqueness of Minimal Polynomial](/theorems/405), $M_\alpha \mid p$, forcing $M_\alpha$ to have distinct linear factors. ($\Leftarrow$) If $M_\alpha$ has distinct linear factors, we decompose $V$ into eigenspaces using Bezout's identity and coprimality, then invoke [Characterisations of Diagonalisability](/theorems/404). [/proofplan] [step:Show that diagonalisability forces $M_\alpha$ to have distinct linear factors] Suppose $\alpha$ is diagonalisable with distinct eigenvalues $\lambda_1, \dots, \lambda_k$. By [Characterisations of Diagonalisability](/theorems/404), $V = \bigoplus_{i=1}^k E_\alpha(\lambda_i)$. Define $p(t) = \prod_{i=1}^k (t - \lambda_i)$. For any $v \in E_\alpha(\lambda_j)$, the factor $(\alpha - \lambda_j\,\mathrm{id})$ sends $v$ to $\mathbf{0}$. Since the factors $(\alpha - \lambda_i\,\mathrm{id})$ commute (they are polynomials in $\alpha$), reorder the product to place $(\alpha - \lambda_j\,\mathrm{id})$ first: \begin{align*} p(\alpha)(v) = \prod_{i \neq j} (\alpha - \lambda_i\,\mathrm{id}) \circ (\alpha - \lambda_j\,\mathrm{id})(v) = \prod_{i \neq j} (\alpha - \lambda_i\,\mathrm{id})(\mathbf{0}) = \mathbf{0}. \end{align*} Every $v \in V$ is a sum of eigenvectors from the direct sum decomposition, and $p(\alpha)$ is linear, so $p(\alpha) = 0$. By [Existence and Uniqueness of Minimal Polynomial](/theorems/405), $M_\alpha(t) \mid p(t)$. Since $p(t)$ is a product of distinct linear factors, every factor of $M_\alpha$ is also a distinct linear factor. [/step] [step:Show that distinct linear factors in $M_\alpha$ force diagonalisability via Bezout decomposition] Suppose $M_\alpha(t) = \prod_{i=1}^k (t - \lambda_i)$ with $\lambda_1, \dots, \lambda_k$ distinct. We prove $V = \bigoplus_{i=1}^k E_\alpha(\lambda_i)$ by induction on $k$. **Base case $k = 1$:** $M_\alpha(t) = t - \lambda_1$ means $\alpha = \lambda_1\,\mathrm{id}$, so $V = E_\alpha(\lambda_1)$. **Inductive step:** Write $M_\alpha(t) = (t - \lambda_k) \cdot g(t)$ where $g(t) = \prod_{i=1}^{k-1}(t - \lambda_i)$. Since $\lambda_1, \dots, \lambda_k$ are distinct, $\gcd(t - \lambda_k, g(t)) = 1$. By Bezout's identity in $\mathbb{F}[t]$, there exist $a(t), b(t) \in \mathbb{F}[t]$ with \begin{align*} a(t)(t - \lambda_k) + b(t)\,g(t) = 1. \end{align*} Evaluating at $\alpha$: $a(\alpha)(\alpha - \lambda_k\,\mathrm{id}) + b(\alpha)\,g(\alpha) = \mathrm{id}$. [guided] The key idea is to use the Bezout identity to construct a direct sum decomposition of $V$. Set $U = \ker(\alpha - \lambda_k\,\mathrm{id}) = E_\alpha(\lambda_k)$ and $W = \ker(g(\alpha))$. The Bezout identity $a(\alpha)(\alpha - \lambda_k\,\mathrm{id}) + b(\alpha)\,g(\alpha) = \mathrm{id}$ will let us decompose every vector as a sum of a piece in $U$ and a piece in $W$. For any $v \in V$, define \begin{align*} w_1 = b(\alpha)\,g(\alpha)(v), \qquad w_2 = a(\alpha)(\alpha - \lambda_k\,\mathrm{id})(v). \end{align*} Then $v = w_1 + w_2$ by the Bezout identity. Why does $w_1 \in U$? Apply $(\alpha - \lambda_k\,\mathrm{id})$ to $w_1$: \begin{align*} (\alpha - \lambda_k\,\mathrm{id})(w_1) = (\alpha - \lambda_k\,\mathrm{id})\,b(\alpha)\,g(\alpha)(v) = b(\alpha)\,M_\alpha(\alpha)(v) = \mathbf{0}, \end{align*} since $(t - \lambda_k)\,g(t) = M_\alpha(t)$ and all polynomial expressions in $\alpha$ commute. So $w_1 \in \ker(\alpha - \lambda_k\,\mathrm{id}) = U$. Why does $w_2 \in W$? Apply $g(\alpha)$ to $w_2$: \begin{align*} g(\alpha)(w_2) = g(\alpha)\,a(\alpha)(\alpha - \lambda_k\,\mathrm{id})(v) = a(\alpha)\,M_\alpha(\alpha)(v) = \mathbf{0}. \end{align*} So $w_2 \in \ker(g(\alpha)) = W$, and therefore $V = U + W$. To show the sum is direct, suppose $v \in U \cap W$, so $(\alpha - \lambda_k\,\mathrm{id})(v) = \mathbf{0}$ and $g(\alpha)(v) = \mathbf{0}$. Applying the Bezout identity: \begin{align*} v = a(\alpha)(\alpha - \lambda_k\,\mathrm{id})(v) + b(\alpha)\,g(\alpha)(v) = \mathbf{0} + \mathbf{0} = \mathbf{0}. \end{align*} So $V = U \oplus W = E_\alpha(\lambda_k) \oplus \ker(g(\alpha))$. The restriction $\alpha|_W$ has minimal polynomial dividing $g(t) = \prod_{i=1}^{k-1}(t - \lambda_i)$. By the inductive hypothesis, $W = \bigoplus_{i=1}^{k-1} E_{\alpha|_W}(\lambda_i) = \bigoplus_{i=1}^{k-1} E_\alpha(\lambda_i)$, and therefore $V = E_\alpha(\lambda_k) \oplus \bigoplus_{i=1}^{k-1} E_\alpha(\lambda_i)$. [/guided] [/step] [step:Conclude diagonalisability from the eigenspace decomposition] The decomposition $V = \bigoplus_{i=1}^k E_\alpha(\lambda_i)$ shows that $\dim V = \sum_{i=1}^k \dim E_\alpha(\lambda_i)$. By [Characterisations of Diagonalisability](/theorems/404), $\alpha$ is diagonalisable. [/step]