MHB Spectral decomposition of compact operators

[Fermat1]

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? Is it simply that we can't decompose the space into a direct sum?

Thanks

Note: this is a duplicate from math stack exchange which has gone unanswered for about a week.

 

[Jhoneine]

Yes, the sequence ($y_{n}$) is the full sequence of eigenvectors of $TT^∗$. In general, if $T$ is a compact operator then it does not have a decomposition into a direct sum. This is what makes the decomposition of a compact operator unsatisfying compared to the decomposition of compact normal operators. The difference between the two is that a normal operator has a decomposition into a direct sum, while a compact operator does not.