# Rank 1 operator on an infinite dimensional vector space number of eigenvalues.

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?

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 $$n$$.