 Solution

We prove independence of eigenvectors for distinct eigenvalues by induction on the number of vectors. A single eigenvector is nonzero and hence independent. Suppose the first $d-1$ are independent, and let ${\sum }_{j=1}^d{c}_j{v}_j=0$. Applying $A-{\lambda }_{d}I$ givesBy the induction hypothesis every coefficient in this shorter relation vanishes. The eigenvalues are distinct, so $c_j=0$ for $j<d$. The original relation becomes $c_d{v}_d=0$, and $v_d\ne 0$ gives $c_d=0$. Thus the entire family of eigenvectors is linearly independent. The argument works over either the real or complex field.