Let $H$ be a non-seperable Hilbert space. Show that every compact operator $T: H \rightarrow H$ has non-separable kernel.
Since $T$ is compact them the image of the unity ball should have compact closure. This doesn't tell me anything about the kernel. Maybe I should assume the kernel is separable and prove somehow that the entire space is separable? I couldn't proceed from here.
Best Answer
Since $\ker T = \ker T^* T$ we may assume that $T$ is additionally self-adjoint. Hence we have $$T = \sum_{k=1}^\infty \lambda_k P_k$$ where $\{P_k\}$ are finite rank pairwise orthogonal projections. In particular, $\operatorname{Ran}T$ has a countable orthonormal basis which can be constructed using orthonormal bases for $\operatorname{Ran}P_k$ and so it follows that $\operatorname{Ran}T$ and hence $\overline{\operatorname{Ran}T}$ are separable. Since $$H = \ker T \oplus \overline{\operatorname{Ran}T}$$ this implies that $\ker T$ must be inseparable since a direct sum of separable spaces is separable.