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No. For example, say $$Ax=\sum\lambda_n\langle x,e_n\rangle e_n,$$where $(e_n)$ are orthomormal and $\lambda_n$ are scalars. Then $A$ is compact if $\lambda_n\to0$, while $A$ is trace class requires $\sum|\lambda_n|<\infty$.

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No. Compact, positive definite operators, for example, are trace class if and only if their eigenvalues are summable

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