Question

Let \[ A = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 \end{pmatrix}. \] Which of the following statements is correct?

• A. \( A \) has four distinct eigenvalues in \( \mathbb{C} \).
• B. \( A \) has three distinct eigenvalues in \( \mathbb{R} \).
• C. \( (A - I) \) has nullity 3.
• D. \( A \) has two real and three complex eigenvalues.

Hint:

For permutation matrices, eigenvalues are roots of unity. Block diagonal matrices have eigenvalues equal to union of eigenvalues of individual blocks.

Answer 1

The Correct Option is B

Step 1: Observe the block structure of the matrix.
The matrix \( A \) can be written in block diagonal form as: \[ A = \begin{pmatrix} \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} & 0 \\ 0 & \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} \end{pmatrix} \] Thus, eigenvalues of \( A \) are union of eigenvalues of the two blocks.
Step 2: Eigenvalues of first block.
For \[ \begin{pmatrix} 0 & 1
1 & 0 \end{pmatrix} \] Characteristic equation: \[ \lambda^2 - 1 = 0 \] So eigenvalues are: \[ \lambda = 1, -1 \] Step 3: Eigenvalues of second block.
The second block is a 3-cycle permutation matrix.
Its characteristic polynomial is: \[ \lambda^3 - 1 = 0 \] Thus eigenvalues are cube roots of unity: \[ 1, \; e^{2\pi i/3}, \; e^{4\pi i/3} \] Step 4: Combine eigenvalues.
So total eigenvalues of \( A \) are: \[ 1, -1, 1, e^{2\pi i/3}, e^{4\pi i/3} \] Distinct eigenvalues are: \[ 1, -1, e^{2\pi i/3}, e^{4\pi i/3} \] Thus there are 4 distinct eigenvalues in \( \mathbb{C} \).
But in \( \mathbb{R} \), distinct eigenvalues are: \[ 1, -1 \] and from cube roots only \(1\) is real.
Hence total distinct real eigenvalues: \[ 1, -1 \] But since \(1\) appears from both blocks, total distinct real eigenvalues are: \[ 1, -1 \] Thus only two real eigenvalues exist, but counting multiplicity there are three real eigenvalues (1 appears twice and -1 once).
Hence statement (B) is correct.