I have a question concerning this question.
In that case, can we say that the orthonormalized eigenvectors of the normal, compact operator T form an orthonormal basis for $H$? If not, under what conditions can we usually guarantee that the eigenvectors of an operator acting on an infinite-dimensional Hilbert space are to form an orthonormal basis? Thanks.
Best Answer
The orthonormalized eigenvectors of a normal compact operator $T$ form an orthonormal basis for $H$:
Since $T$ is normal, eigenvectors of distinct eigenvalues are orthogonal. For the same eigenvalue, they can be orthonormalized.
Since $T$ is compact, there is a countable set of eigenvectors with non-zero eigenvalues. Each eigenvalue has a finite number of independent eigenvectors, hence the non-zero eigenvalues give a countable set of orthonormal eigenvectors.
The answer to the referenced question shows that the perpendicular space is then the eigenspace of the zero eigenvalue (unless it is trivial). This space need not be finite dimensional or even have a countable basis. However any Hilbert space, here $\ker T$, has an orthonormal basis of some cardinality.