 Solution

Apply $A$ to the stated linear combination and use each eigenvector equation:Subtract ${\lambda }_{k+1}$ times the original relation. This givesThe assumed linear independence implies ${\alpha }_i({\lambda }_i-{\lambda }_{k+1})=0$ separately for every $i$. Since the eigenvector $z_{k+1}$ is nonzero, not all ${\alpha }_i$ can vanish. Choose any $i$ with ${\alpha }_i\ne 0$; division by that coefficient gives ${\lambda }_i={\lambda }_{k+1}$. In fact every nonzero coefficient is attached to that same eigenvalue.