- #1
Ricky2357
- 23
- 0
Suppose T: X -> Y and S: Y -> Z , X,Y,Z normed spaces , are bounded linear operators. Is there an example where T and S are not the zero operators but SoT (composition) is the zero operator?
Ricky2357 said:R with absolute value is simple enough
Ricky2357 said:It is enough to find two nonzero matrices such that their product is zero [note correction].
Ricky2357 said:Why is that? A linear operator from R into R of the form Tx=k*x is bounded.
(absolute value as norm)
Ricky2357 said:Second, the image (range) of a bounded linear operator is not always a bounded set. A bounded linear operator is called bounded because it maps bounded sets into bounded sets.
A bounded linear operator is a type of function that maps between two vector spaces, typically in functional analysis. It is linear in that it preserves addition and scalar multiplication, and it is bounded in that it does not grow too quickly in magnitude as the input increases.
A general linear operator does not have the restriction of being bounded, meaning it can grow or shrink arbitrarily as the input changes. A bounded linear operator, on the other hand, is limited in its growth and has a maximum rate at which it can increase.
Some common examples of bounded linear operators include differentiation and integration operators, as well as many common matrix operations such as matrix multiplication and transposition.
Boundedness is important in functional analysis because it allows us to study linear operators in a more rigorous and consistent way. It also allows us to make statements about the behavior and properties of these operators with more certainty and precision.
In general, the boundedness of a linear operator is related to the size and behavior of its eigenvalues. A bounded linear operator will have eigenvalues that are finite and do not grow too quickly, while an unbounded operator may have eigenvalues that grow without bound. Additionally, the boundedness of an operator can be characterized by the size of its largest eigenvalue.