Chapter 13: Problem 26

Show that every complex number \(\lambda\) in the spectrum of a unitary operator has \(|\lambda|=1\).

Short Answer

Expert verified

For a given unitary operator, we utilize its properties and the eigenvalue equation to prove that every complex number in the spectrum of the operator has an absolute value equal to 1. This is done by first obtaining an equation related to the absolute value squared of the eigenvalue, and then simplifying that equation to show that the absolute value of the eigenvalue is indeed 1.

Step by step solution

Understand the properties of a unitary operator

A unitary operator \( U \) is an operator that satisfies the condition \( UU^* = U^*U = I \), where \( U^* \) is the conjugate transpose of \( U \), and \( I \) is the identity matrix. This property will be crucial in proving the statement.

Understand the definition of the spectrum of an operator

The spectrum of an operator is essentially the set of its eigenvalues. An eigenvalue \(\lambda\) associated with operator \(A\) satisfies the equation \(Ax = \lambda x\), where \(x\) is an eigenvector corresponding to \(\lambda\). In this case, we're asked to prove that the absolute value of each eigenvalue \(\lambda\) of a unitary operator is 1.

Prove the statement

For any eigenvalue \(\lambda\), we have that \(Ux=\lambda x\) for some non-zero vector \(x\). Taking the conjugate transpose of the equation yields \((Ux)^*=\lambda^* x^*\). By further simplifying this using the properties of a unitary operator and the fact that \(\lambda\) is an eigenvalue, we have that \(x^*U^*=\lambda^* x^*\). This would lead us to show that \(x^*U^*Ux=x^*Ix = x^*x\); also, on simplifying further we get \(\lambda \lambda^* x^*x = x^*x\). Hence, we have \(|\lambda|^2 = 1\). Therefore, \(|\lambda| = 1\), thus proving the statement.