Normal operators with real eigenvalues are self-adjoint

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SUMMARY

A normal operator with real eigenvalues is self-adjoint, as established through the diagonalization process. Given a normal operator A, it can be expressed as A = UDU*, where D is a diagonal matrix containing real eigenvalues and U is a unitary matrix. The proof relies on the property that if D is real, then D = D*, leading to the conclusion that A = A*. The discussion highlights the importance of understanding the relationship between normal operators and their eigenvalues in the context of linear algebra.

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  • Understanding of normal operators in linear algebra
  • Familiarity with unitary matrices and their properties
  • Knowledge of diagonalization of matrices
  • Concept of self-adjoint operators
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  • Study the properties of normal operators in detail
  • Learn about the spectral theorem for normal operators
  • Explore the implications of self-adjoint operators in quantum mechanics
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Prove that a normal operator with real eigenvalues is self-adjoint


Seems like a simple proof, but I can't seem to get it.

My attempt: We know that a normal operator can be diagonalized, and has a complete orthonormal set of eigenvectors.

Let A be normal. Then A= UDU* for some diagonal matrix D and unitary U. Also, A*=U*D*U

Since D is the diagonal matrix of the eigen values of A, D is real, and thus D=D*.

Thus D=U*AU = UA*U*.

Then, I just get stuck on A=U²A*(U*)².
 
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You seem to be using the substitution
(ST)* --> S*T*,​
but the left hand side is not equal to the righthand side; making this substitution is not a valid deduction.
 

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