Question

Let \(n>1\) be an integer. Consider the following two statements for an arbitrary \(n \times n\) matrix \(A\) with complex entries. I. If \(A^k = I_n\) for some integer \(k \ge 1\), then all the eigenvalues of \(A\) are \(k^{\text{th}}\) roots of unity.
II. If, for some integer \(k \ge 1\), all the eigenvalues of \(A\) are \(k^{\text{th}}\) roots of unity, then \(A^k = I_n.\)
Then

• A. both I and II are TRUE.
• B. I is TRUE but II is FALSE.
• C. I is FALSE but II is TRUE.
• D. neither I nor II is TRUE.

Hint:

The condition on eigenvalues ensures behavior of diagonalizable matrices only. Non-diagonalizable matrices can break such implications.

Answer 1

The Correct Option is B

Step 1: Analyze Statement I.
If \(A^k = I_n\), then by the spectral theorem, every eigenvalue \(\lambda\) satisfies \(\lambda^k = 1\). Hence, all eigenvalues are \(k^{\text{th}}\) roots of unity. So (I) is TRUE.
Step 2: Analyze Statement II.
If all eigenvalues of \(A\) are \(k^{\text{th}}\) roots of unity, it does \emph{not} imply \(A^k = I_n\), since \(A\) may not be diagonalizable. For example, \[ A = \begin{pmatrix} 1 & 1
0 & 1 \end{pmatrix} \] has all eigenvalues = 1 (a 1st root of unity), but \(A^k \ne I_n.\) Hence, (II) is FALSE.
Step 3: Conclusion.
Statement I is TRUE, Statement II is FALSE. Hence option (B).