= Solution
Apply $A$ to the stated <linear combination> and use each <eigenvector> equation:
$$
\lambda_{k+1}z_{k+1}=Az_{k+1}=\sum_{i=1}^k\alpha_iAz_i=\sum_{i=1}^k\alpha_i\lambda_i z_i.
$$
Subtract $\lambda_{k+1}$ times the original relation. This gives
$$
\sum_{i=1}^k\alpha_i(\lambda_i-\lambda_{k+1})z_i=0.
$$
The 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 $\alpha_i\ne0$; division by that coefficient gives \b[$\lambda_i=\lambda_{k+1}$]. In fact every nonzero coefficient is attached to that same <eigenvalue>.
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