 Solution

We use induction on $d$. SupposeApplying $f-{\alpha }_{d}I$ givesBy induction, every $c_i({\alpha }_i-{\alpha }_d)=0$ for $i<d$. The eigenvalues are distinct, so $c_i=0$ for $i<d$, and the original relation then gives $c_d=0$. Thus eigenvectors with distinct eigenvalues are linearly independent.