Spectral decomposition of compact operators

Question

Let T be a compact operator on an infinite dimensional hilbert space H. I am proving the theorem which says that $Tx=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,x_{n}\rangle y_{n}$ where ($x_{n}$) is an orthonormal sequence consisting of the eigenvectors of $|T|=(T^*T)^{0.5}$, (${\lambda}_{n}$) is the corresponding sequence of eigenvalues, and ($y_{n}$) is an orthonormal sequence of, each $y_{n}$ is an eigenvector of $TT^*$.

I have two questions. Is ($y_{n}$) the full sequence of eigenvectors of $TT^*$? I would have thought the answer was yes because the square root of an operator has the same eigenvectors as the operator and an operator has the same 'number' of eigenvectors as it's adjoint.

Secondly, what makes this decomposition so unsatisfying compared to the decomposition of compact normal operators?

This question wasn't answered on stack exchange so I thought it might be more appropriate here. Thanks

Answer 1

What you describe is the so-called singular value decomposition (SVD) of the compact operator $T$, I infer. In this case, $(y_n)$ is the full sequence of eigenvectors of $TT^*$ if and only if zero is not an eigenvalue of $T$.

As for your second question, it depends on what you call "unsatisfying". The singular value decomposition of $T$ is extremely useful in many aspects - for instance, it provides an efficient method to compute the pseudo-inverse of $T$. Supposing that by "decomposition of compact normal operators" you mean "spectral decomposition", the singular value decomposition is, in a sense, the best you can expect from a not necessarily normal, compact operator $T$, since at least in finite dimensions $T$ has a spectral decomposition (i.e. it is diagonalizable with respect to a single basis) if and only if it is normal with respect to some scalar product. See also the Wikipedia entry for SVD for more details and references.

Answer 2

This a comment rather than an example but I don't have this option. An instructive example of what can go on is provided by the weighted shift operator $$(x_1,x_2,x_3,\dots)\mapsto (0,x_1,\frac 1 2 x_2,\frac 1 3 x_3,\dots).$$ (By the way, I think that the requisite condition is that the range of the operator be dense---clearly not fulfilled here).