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Does a rank 1 bounded operator from $\mathscr{K}:L^2([0,1])\to L^2([0,1])$ have at most 1 non-zero eigenvalue? The reason this is not obvious to me is that $L^2([0,1])$ is infinite dimensional. In general, is rank a bound on the number of eigenvalues?

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Yes, in general, this is true: the rank is an upper bound on the number of non-zero eigenvalues.

If we have a linear operator $T : X \to X$, where $X$ is not necessarily finite-dimensional, then saying $T$ has finite rank $n$ means that $\operatorname{Im} T$ is has finite dimension $n$. Note also that, if $v \in X$ is an eigenvector for eigenvalue $\lambda \neq 0$, then $$v = \lambda^{-1}\lambda v = T(\lambda^{-1}v) \in \operatorname{Im} T.$$ Finally, as we always have, eigenvectors for distinct eigenvalues are linearly independent. The standard proof works in general, and does not require finite-dimensionality (you can find a proof here). This means that we can only fit at most $n$ such eigenvalues into the $\operatorname{Im}(T)$ and still keep the dimension $n$.

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