Question

Let \( n \ge 2 \) be an integer. Let \( A : \mathbb{C}^n \to \mathbb{C}^n \) be the linear transformation defined by \[ A(z_1, z_2, \ldots, z_n) = (z_n, z_1, z_2, \ldots, z_{n-1}). \] Which one of the following statements is true for every \( n \ge 2 \)?

• A. \(A\) is nilpotent.
• B. All eigenvalues of \(A\) are of modulus \(1.\)
• C. Every eigenvalue of \(A\) is either \(0\) or \(1.\)
• D. \(A\) is singular.

Hint:

The cyclic shift matrix satisfies \(A^n = I\), so its eigenvalues are exactly the \(n\)-th roots of unity — all having modulus 1.

Answer 1

The Correct Option is B

Step 1: Identify the matrix form of \(A.\)
\(A\) is the cyclic right-shift operator on \(\mathbb{C}^n\). Its matrix representation is \[ A = \begin{pmatrix} 0 & 0 & 0 & \cdots & 1
1 & 0 & 0 & \cdots & 0
0 & 1 & 0 & \cdots & 0
\vdots & & \ddots & \ddots & \vdots
0 & 0 & \cdots & 1 & 0 \end{pmatrix}. \]
Step 2: Characteristic polynomial and eigenvalues.
Since \(A^n = I\), all eigenvalues \(\lambda\) satisfy \(\lambda^n = 1\). Hence, the eigenvalues are the \(n\)-th roots of unity: \[ \lambda_k = e^{2\pi i k/n}, \quad k = 0,1,2,\ldots,n-1. \]
Step 3: Modulus of eigenvalues.
Each eigenvalue satisfies \(|\lambda_k| = 1\). Therefore, all eigenvalues of \(A\) lie on the unit circle.
Step 4: Conclusion.
Thus, (B) is correct.