[proofplan] We prove by induction on $k$ that if $v_i \in E_\alpha(\lambda_i)$ for $i = 1, \dots, k$ and $v_1 + \cdots + v_k = \mathbf{0}$, then every $v_i = \mathbf{0}$. The inductive step applies $\alpha$ to a dependence relation and subtracts $\lambda_k$ times the original to eliminate the last eigenspace, reducing to $k-1$ eigenvalues. [/proofplan] [step:Handle the base case $k = 1$] A single eigenvector $v_1 \in E_\alpha(\lambda_1)$ is non-zero by definition. If $v_1 = \mathbf{0}$, the sum is trivially zero, so the directness condition holds vacuously. [/step] [step:Apply $\alpha$ to a dependence relation and subtract to eliminate $v_k$] Assume the result holds for $k - 1$ distinct eigenvalues. Let $\lambda_1, \dots, \lambda_k$ be distinct eigenvalues, and suppose \begin{align*} v_1 + v_2 + \cdots + v_k = \mathbf{0}, \quad v_i \in E_\alpha(\lambda_i). \end{align*} Apply $\alpha$ to both sides. Since $\alpha(v_i) = \lambda_i v_i$ for each $i$: \begin{align*} \lambda_1 v_1 + \lambda_2 v_2 + \cdots + \lambda_k v_k = \mathbf{0}. \end{align*} Subtract $\lambda_k$ times the original relation: \begin{align*} (\lambda_1 - \lambda_k)v_1 + (\lambda_2 - \lambda_k)v_2 + \cdots + (\lambda_{k-1} - \lambda_k)v_{k-1} = \mathbf{0}. \end{align*} Each vector $(\lambda_i - \lambda_k)v_i$ lies in $E_\alpha(\lambda_i)$ since eigenspaces are subspaces. By the inductive hypothesis applied to the $k - 1$ distinct eigenvalues $\lambda_1, \dots, \lambda_{k-1}$, the sum $E_\alpha(\lambda_1) + \cdots + E_\alpha(\lambda_{k-1})$ is direct. Therefore $(\lambda_i - \lambda_k)v_i = \mathbf{0}$ for each $i \in \{1, \dots, k-1\}$. [guided] The strategy is to produce a relation involving only $k - 1$ eigenspaces, then invoke the inductive hypothesis. We have two relations: the original $\sum_{i=1}^k v_i = \mathbf{0}$ and the result of applying $\alpha$, namely $\sum_{i=1}^k \lambda_i v_i = \mathbf{0}$. Subtracting $\lambda_k$ times the first from the second eliminates the $v_k$ term entirely: \begin{align*} \sum_{i=1}^{k-1} (\lambda_i - \lambda_k)v_i = \mathbf{0}. \end{align*} Why does this still involve eigenvectors for the respective eigenspaces? Each $(\lambda_i - \lambda_k)v_i$ is a scalar multiple of $v_i \in E_\alpha(\lambda_i)$, and eigenspaces are subspaces (closed under scalar multiplication), so $(\lambda_i - \lambda_k)v_i \in E_\alpha(\lambda_i)$. The inductive hypothesis applies because we now have a dependence relation among elements of $k - 1$ distinct eigenspaces, giving $(\lambda_i - \lambda_k)v_i = \mathbf{0}$ for each $i = 1, \dots, k-1$. [/guided] [/step] [step:Conclude that every $v_i = \mathbf{0}$] Since $\lambda_i \neq \lambda_k$ for $i \in \{1, \dots, k-1\}$, the scalar $\lambda_i - \lambda_k \neq 0$ in $\mathbb{F}$. From $(\lambda_i - \lambda_k)v_i = \mathbf{0}$ and $\lambda_i - \lambda_k \neq 0$, we conclude $v_i = \mathbf{0}$ for $i = 1, \dots, k-1$. Substituting back into the original relation $v_1 + \cdots + v_k = \mathbf{0}$ gives $v_k = \mathbf{0}$. Every $v_i = \mathbf{0}$, so the sum $E_\alpha(\lambda_1) + \cdots + E_\alpha(\lambda_k)$ is direct. [/step]