= Solution
Let $Av_1=\lambda_1v_1$ and $Av_2=\lambda_2v_2$, with both vectors nonzero and $\lambda_1\ne\lambda_2$. If $c_1v_1+c_2v_2=0$, apply $A-\lambda_2I$ to get $c_1(\lambda_1-\lambda_2)v_1=0$. Hence $c_1=0$, and then $c_2=0$.
For the general <independence of eigenvectors for distinct eigenvalues>, use induction. In a relation $\sum_{j=1}^n c_jv_j=0$, applying $A-\lambda_nI$ gives $\sum_{j=1}^{n-1}c_j(\lambda_j-\lambda_n)v_j=0$. Independence of the first $n-1$ vectors makes each coefficient vanish. Distinctness of the eigenvalues then gives $c_1=\cdots=c_{n-1}=0$, and the original relation gives $c_n=0$. No Hermitian assumption is required.
Back to article page
