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.
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.
To prove that the orthogonal projections $P_k$ in $T$'s spectral decomposition are finite-rank operators (part of the "spectral theorem"), you actually need to know that $\overline{\operatorname{ran}(T)}$ is separable. Here is another proof of this fact without using the spectral theorem.
If we assume the existence of a "compact normal operator $T$ on a Hilbert space $\mathcal{H}$", then automatically $\mathcal{H} = \operatorname{ker}(T) \oplus \overline{\operatorname{ran}(T)}$, where $T$ vanishes on $\operatorname{ker}(T)$, and $\overline{\operatorname{ran}(T)}$ is separable. Therefore we can reduce the proof of the spectral theorem to the case where $\mathcal{H}$ is separable.
(If anyone thinks this should rather be a comment: I'll be happy if you can my answer to the comments above, sorry I don't have enough reputation.)
