Eigenvectors of distinct eigenvalues are independent

We argue by induction on $k$. For $k = 1$ the claim is trivial: a single nonzero vector is automatically linearly independent.

Assume the result holds for $k - 1$ distinct eigenvalues, and suppose toward contradiction that $\mathbf{v}_1, \dots, \mathbf{v}_k$ are linearly dependent. Then there is a relation $c_1 \mathbf{v}_1 + \cdots + c_k \mathbf{v}_k = \mathbf{0}$ with not all $c_i$ zero.

Apply $A$ to both sides: since $A\mathbf{v}_i = \lambda_i \mathbf{v}_i$, this gives $c_1 \lambda_1 \mathbf{v}_1 + \cdots + c_k \lambda_k \mathbf{v}_k = \mathbf{0}$. Now multiply the original relation by $\lambda_k$ and subtract it from this new equation; the terms with $\mathbf{v}_k$ cancel exactly, leaving $c_1(\lambda_1 - \lambda_k)\mathbf{v}_1 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k)\mathbf{v}_{k-1} = \mathbf{0}$.

By the induction hypothesis, $\mathbf{v}_1, \dots, \mathbf{v}_{k-1}$ are linearly independent, so every coefficient in this last relation must vanish: $c_i(\lambda_i - \lambda_k) = 0$ for $i < k$. Since the eigenvalues are pairwise distinct, $\lambda_i - \lambda_k \neq 0$, forcing $c_i = 0$ for all $i < k$.

Substituting back into the original relation leaves $c_k \mathbf{v}_k = \mathbf{0}$; since $\mathbf{v}_k \neq \mathbf{0}$ by definition of eigenvector, $c_k = 0$ as well. Every coefficient is zero, contradicting our assumption that not all $c_i$ vanish. Hence no such dependence relation exists, and the eigenvectors are linearly independent.