Solving Self-Adjoint Problem in Complex Inner Product Space

  • Thread starter redyelloworange
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In summary: T = T_1 + T_2 is trivially self adjoint, T* = T_1* + T_2* = T_1 + T_2 = TFor b) we haveU_1 = T_1 + iT_2U_1* = T_1* + i T_2* = T_1 + iT_2 = U_1U_2 = T_1- iT_2U_2* = T_1* - i T_2* = T_1 - iT_2 = U_2For c), one direction (normal => T_1 T_2 = T_2T_1)T normal
  • #1
redyelloworange
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Homework Statement


Let V be a complex inner product space, and let T be a linear operator on V.
Define

T_1 = 1/2 (T + T*) and T_2 = (1/2i)(T – T*)

a) Prove that T_1 and T_2 are self-adjoint and that T = T_1 + T_2
b) Suppose also that T = U_1 + U_2, where U_1 and U_2 are self-adjoint. Prove that U_1 = T_1 and U_2 = U_2.
c) Prove that T is normal iff (T_1)(T_2) = (T_2)(T_1)

Homework Equations


Self-adjoint: T = T*
Normal: TT* = T*T

The Attempt at a Solution



(a) was very easy. However, I cannot get started on (b) at all.
For (c)
One direction: assume (T_1)(T_2) = (T_2)(T_1), show T is normal

Substitute:

(1/2 (T + T*))((1/2i)(T – T*)) = ((1/2i)(T – T*))( 1/2 (T + T*))

(1/4i)(T^2 – T* ^2) = (1/4i)(T^2 – T*^2)

I’m not sure what to do now, or if it’s even the right direction.

Thanks for your help!
 
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  • #2
For b, you must mean

"Suppose also that T = U_1 + i U_2, where U_1 and U_2 are self-adjoint. Prove that U_1 = T_1 and U_2 = T_2."

This shouldn't be too hard.
 
  • #3
redyelloworange said:

Homework Statement


Let V be a complex inner product space, and let T be a linear operator on V.
Define

T_1 = 1/2 (T + T*) and T_2 = (1/2i)(T – T*)

a) Prove that T_1 and T_2 are self-adjoint and that T = T_1 + T_2
Are you sure of this? It doesn't look to me like T_2 is self adjoint and it is easy to see that T_1+ T_2 is NOT T! Did you mean T= T_1+ i T_2? This is a lot like breaking ex into cosine and sine but, again, T_2 does not satisfy <T_2 u, v>= <u, T_2 v>. It satisfies <T_2u, v>= -<u, T_2 v>.

b) Suppose also that T = U_1 + U_2, where U_1 and U_2 are self-adjoint. Prove that U_1 = T_1 and U_2 = U_2.
As StatusX said, this should be U_2= T_2.


c) Prove that T is normal iff (T_1)(T_2) = (T_2)(T_1)

Homework Equations


Self-adjoint: T = T*
Normal: TT* = T*T

The Attempt at a Solution



(a) was very easy. However, I cannot get started on (b) at all.
For (c)
One direction: assume (T_1)(T_2) = (T_2)(T_1), show T is normal

Substitute:

(1/2 (T + T*))((1/2i)(T – T*)) = ((1/2i)(T – T*))( 1/2 (T + T*))

(1/4i)(T^2 – T* ^2) = (1/4i)(T^2 – T*^2)

I’m not sure what to do now, or if it’s even the right direction.

Thanks for your help!
 
  • #4
Sorry about that,

It should be

T = T_1 + iT_2

T_2 is self adoint: T_2* = (-1/2i)(T*-T) = (1/2i)(T-T*) = T_2
 

1. What is a self-adjoint problem in a complex inner product space?

A self-adjoint problem in a complex inner product space refers to a mathematical problem where the solution is equivalent to its adjoint, or Hermitian conjugate. This means that when the problem is solved, the result and its transpose are equal.

2. How is a self-adjoint problem solved in a complex inner product space?

To solve a self-adjoint problem in a complex inner product space, one must use the concept of Hermitian operators. These are operators that satisfy the condition that the adjoint of the operator is equal to the operator itself.

3. What are the applications of solving self-adjoint problems in complex inner product spaces?

Solving self-adjoint problems in complex inner product spaces has various applications in physics, engineering, and mathematics. Some examples include solving partial differential equations, finding eigenvalues and eigenvectors, and studying quantum mechanics.

4. What are the key properties of self-adjoint problems in complex inner product spaces?

The key properties of self-adjoint problems in complex inner product spaces include symmetry, orthogonality, and completeness. These properties play a crucial role in solving such problems and understanding their solutions.

5. Can self-adjoint problems in complex inner product spaces have multiple solutions?

Yes, self-adjoint problems in complex inner product spaces can have multiple solutions. This is because the solutions are not unique and can be multiplied by a complex constant without affecting the adjoint condition. However, these solutions are still considered equivalent and provide the same information about the problem.

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