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Chapter 13: Problem 57

Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal.

Short Answer

Expert verified
In conclusion, we proved that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal by showing that they satisfy the condition AA* = A*A. For self-adjoint operators, we used the fact that A = A*. For skew-adjoint operators, we used A* = -A and for unitary (orthogonal) operators, we used AA* = A*A = I, thus satisfying the definition of normal operators.

Step by step solution

01

Self-adjoint operator

For a self-adjoint operator A, we have A = A*. We need to prove that AA* = A*A. Since A = A*, we have: AA* = A*A Thus, a self-adjoint operator is normal.
02

Skew-adjoint operator

For a skew-adjoint operator A, we have A* = -A. We need to prove that AA* = A*A. When looking at the operator product, we get: AA* = A(-A) = -AA And, for the adjoint product: A*A = (-A)*A = -A*A, since (-A)* = -A* for a skew-adjoint operator. So, AA* = -AA = A*A, and a skew-adjoint operator is normal.
03

Unitary (orthogonal) operator

For a unitary (orthogonal) operator A, we have AA* = A*A = I. We need to prove that AA* = A*A. Since the condition for a unitary (orthogonal) operator is given by: AA* = A*A = I We directly see that the product of the operator A with its adjoint is equal to the product of its adjoint with the operator, which is the definition of a normal operator. Thus, a unitary (orthogonal) operator is normal. In conclusion, we have shown that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are all normal operators.

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