 Solution

Normalize the mutually orthogonal eigenvectors so that $(x_j)$ is an orthonormal sequence. If ${\lambda }_j$ did not tend to zero, some subsequence would satisfy $|{\lambda }_j|\ge \epsilon >0$. For distinct indices in this subsequence, orthogonality givesThus $(T{x}_j)$ would have no Cauchy subsequence, contradicting compactness of $T$. Therefore the orthogonal eigenvectors of a compact operator satisfy