Spectral decomposition of compact operators

In summary: This is due to the fact that the operator $|T|$ does not have a decomposition into a direct sum, and therefore the eigenvectors of $|T|$ cannot be used to construct a direct sum decomposition for $T$. In summary, the theorem being proved states that a compact operator on an infinite dimensional Hilbert space can be decomposed into a sum of eigenvalues and eigenvectors, where the eigenvectors are orthonormal sequences of both $|T|$ and $TT^*$. This decomposition is not as satisfying as the decomposition for compact normal operators because it cannot be written as a direct sum.
  • #1
Fermat1
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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.
 
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  • #2
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.
 

FAQ: Spectral decomposition of compact operators

What is spectral decomposition of compact operators?

Spectral decomposition of compact operators is a mathematical technique used to break down a compact operator into simpler components, making it easier to analyze and understand its behavior. This method is often used in functional analysis and linear algebra.

How does spectral decomposition help in understanding compact operators?

Spectral decomposition helps in understanding compact operators by breaking them down into their eigenvalues and eigenvectors. This allows us to study the behavior of the operator on each eigenspace separately, providing insights into the overall behavior of the operator.

What is the significance of compact operators in mathematics?

Compact operators are important in mathematics because they are used to represent linear transformations between infinite-dimensional vector spaces. They also have important applications in functional analysis, differential equations, and quantum mechanics.

Can spectral decomposition be applied to non-compact operators?

No, spectral decomposition is only applicable to compact operators. Non-compact operators do not have a discrete spectrum, which is a necessary condition for spectral decomposition.

Are there any limitations or drawbacks of spectral decomposition?

One limitation of spectral decomposition is that it can only be applied to compact operators. Additionally, it may not always be possible to find a complete set of eigenvalues and eigenvectors for a given compact operator, making it difficult to fully decompose the operator.

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