If A*A is a compact operator, is A compact?

Thread starter elina
Start date May 25, 2009
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Compact Operator

In summary, the compactness principle states that if A*A is compact, then A is compact. A compact operator is a linear operator on a Banach space that maps bounded sets to relatively compact sets. If A*A is compact, it implies that A is a Hilbert-Schmidt operator and allows for the use of the spectral theorem. It is possible for a non-compact operator to have a compact square. Other conditions for A to be compact include being a bounded operator, having a precompact image of the unit ball, and having finite rank.

May 25, 2009

elina

Hi, if I know that A is continuous and linear and A*A is compact, where A* is the adjungate, how will I show that A has to be compact?

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May 25, 2009

gel

You could try using the http://en.wikipedia.org/wiki/Polar_decomposition" .

Last edited by a moderator: May 4, 2017

Related to If A*A is a compact operator, is A compact?

1. Is it true that if A*A is compact, then A is compact?

Yes, if A*A is compact, then A is compact. This is known as the "compactness principle" and is a result of the fact that the composition of two compact operators is also compact.

2. What is a compact operator?

A compact operator is a type of linear operator on a Banach space (a complete normed vector space) that maps bounded sets to relatively compact sets. In other words, it maps bounded sets to sets that have compact closures.

3. What is the significance of A*A being compact?

If A*A is compact, it implies that A is a Hilbert-Schmidt operator, which has important properties in functional analysis. It also allows for the use of the spectral theorem, which is a powerful tool in the study of compact operators.

4. Can a non-compact operator have a compact square?

Yes, it is possible for a non-compact operator to have a compact square. For example, the operator A = [1, 0; 0, 0] is non-compact, but A*A = [1, 0; 0, 0] is compact.

5. Are there any other conditions for A to be compact besides A*A being compact?

Yes, there are other conditions for A to be compact. One such condition is that A must be a bounded operator. Another condition is that the image of the unit ball under A must be precompact, meaning its closure is compact. Additionally, A must have finite rank in order to be compact.