= 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 $\sum_{j=1}^d c_jv_j=0$. Applying $A-\lambda_dI$ gives
$$
\sum_{j=1}^{d-1}c_j(\lambda_j-\lambda_d)v_j=0.
$$
By 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 $c_dv_d=0$, and $v_d\ne0$ gives $c_d=0$. Thus \b[the entire family of <eigenvectors> is <linearly independent>]. The argument works over either the real or complex field.
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