HallsofIvy said:(AB)*= B*A* and (A*)*= A.
The adjoint of T*T-TT* is (T*T- TT*)*=(T*T)*- (TT*)*= (T*)(T**)- (T**)T*. And the adjoint is 'dual'- that is, T**= T so that becomes T*T- TT* again.
An operator is considered normal if it commutes with its adjoint operator. In other words, the operator and its adjoint can be applied in any order without changing the result.
This notation is representing the norm of an operator applied to a vector. The double vertical bars indicate the norm, and the "v" and "*v" represent two different vectors that the operator is being applied to.
One way to prove this statement is by using the definition of a normal operator and the properties of the norm. By definition, a normal operator commutes with its adjoint, which means that ||Tv||=||T*v||. On the other hand, if ||Tv||=||T*v||, we can show that the operator commutes with its adjoint, thus proving that it is normal.
Yes, a simple proof would involve using the definition of a normal operator and the properties of the norm, as mentioned in the previous answer. You can also use the fact that an operator is normal if and only if its eigenvectors form an orthonormal basis, and then use the properties of the norm to show that ||Tv||=||T*v|| for all eigenvectors.
Yes, this statement has many applications in various fields such as physics, engineering, and computer science. For example, in quantum mechanics, normal operators are used to represent physical observables, and this statement helps to prove their properties. In signal processing, normal operators are used to analyze signals, and this statement helps to simplify calculations and proofs.