Question

Which of the following can’t be an eigenvalue of any Unitary matrix?

• A. \( \dfrac{3}{2} + \dfrac{i}{2} \)
• B. \( \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i \)
• C. \( \dfrac{\sqrt{7} + \sqrt{2}i}{3} \)
• D. \( i \)

Hint:

Eigenvalues of unitary matrices lie on the unit circle \( \Rightarrow |\lambda| = 1 \).

Answer 1

The Correct Option is A

A Unitary matrix \( U \) satisfies \( U^*U = I \), where \( U^* \) is the conjugate transpose, and \( I \) is the identity matrix. The eigenvalues \( \lambda \) of a Unitary matrix have the property \( |\lambda| = 1 \), meaning they lie on the unit circle in the complex plane.

Let's verify which eigenvalue does not satisfy this condition:

• \(\lambda = \dfrac{3}{2} + \dfrac{i}{2}\)
  The modulus is \(|\lambda| = \sqrt{\left(\dfrac{3}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2} = \sqrt{\dfrac{9}{4} + \dfrac{1}{4}} = \sqrt{\dfrac{10}{4}} = \sqrt{\dfrac{5}{2}}\). Thus, \(|\lambda| \neq 1\).
• \(\lambda = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\)
  The modulus is \(|\lambda| = \sqrt{\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2} = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} = \sqrt{1} = 1\).
• \(\lambda = \dfrac{\sqrt{7} + \sqrt{2}i}{3}\)
  The modulus is \(|\lambda| = \sqrt{\left(\dfrac{\sqrt{7}}{3}\right)^2 + \left(\dfrac{\sqrt{2}}{3}\right)^2} = \sqrt{\dfrac{7}{9} + \dfrac{2}{9}} = \sqrt{\dfrac{9}{9}} = 1\).
• \(\lambda = i\)
  The modulus is \(|i| = \sqrt{0^2 + 1^2} = 1\).

Thus, the eigenvalue \( \dfrac{3}{2} + \dfrac{i}{2} \) does not have a modulus of 1 and, therefore, cannot be an eigenvalue of any Unitary matrix.