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elina
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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?
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.
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.
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.
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.
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.