Let $\mathcal{R}$ denote the range space and $\mathcal{N}$ the null space of a linear transformation. We write $$ \mathcal{R}(t^n) = \mathcal{R}(t^{n+1}) = \dots = \mathcal{R}_{\infty}(t)$$ and likewise $$ \mathcal{N}(t^n) = \mathcal{N}(t^{n+1}) = \dots = \mathcal{N}_{\infty}(t).$$
In the book of Jim Hefferon - on his way to Jordan Form - he proves that $$\mathcal{N}_{\infty}(t-\lambda_i) \cap \mathcal{N}_{\infty}(t-\lambda_j) = \{0 \}$$ for eigenvalues $\lambda_i \neq \lambda_j$ for a linear endomorphism $t$ on a finite $\mathbb{C}$-vector space $V$.
He later notes that $$ \mathcal{R}_\infty(t-\lambda_1) \cap \dots \cap \mathcal{R}_\infty(t-\lambda_k) = \{0 \}$$ where $\lambda_1, \dots, \lambda_k$ are all eigenvalues of $t$ follows from the fact above. However, I cannot seem to see how it follows. What am I missing?