Suppose $c_1v_1 + \cdots + c_rv_r = \mathbf{0}$ is a nontrivial relation, and let $k$ be the number of nonzero coefficients (so $k \geq 1$). Among all nontrivial relations, choose one with the *fewest* nonzero terms. Relabelling, assume $c_1, \ldots, c_k \neq 0$ and $c_{k+1} = \cdots = c_r = 0$, so $c_1v_1 + \cdots + c_kv_k = \mathbf{0}$. Apply $A$: $c_1\lambda_1v_1 + \cdots + c_k\lambda_kv_k = \mathbf{0}$. Subtract $\lambda_k$ times the original relation: $c_1(\lambda_1 - \lambda_k)v_1 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k)v_{k-1} = \mathbf{0}$. This is a relation with at most $k - 1$ nonzero terms (since $\lambda_i - \lambda_k \neq 0$ for $i < k$ by distinctness, and $c_i \neq 0$). By minimality of $k$, this relation must be trivial, so $k - 1 = 0$, meaning the original relation had only one term: $c_1v_1 = \mathbf{0}$. Since $v_1 \neq \mathbf{0}$, this forces $c_1 = 0$, contradicting $c_1 \neq 0$.