Question

Let \( A \) be a \( 2 \times 2 \) matrix as given:

\[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
What are the eigenvalues of the matrix \( A^{13} \)?

• A. 1, −1
• B. \( 2\sqrt{2}, -2\sqrt{2} \)
• C. \( 4\sqrt{2}, -4\sqrt{2} \)
• D. \( 64\sqrt{2}, -64\sqrt{2} \)

Hint:

When raising a matrix to a power, the eigenvalues of the resulting matrix are the eigenvalues of the original matrix raised to that power. This property is useful for calculating eigenvalues of powers of matrices efficiently.

Answer 1

The Correct Option is D

To solve for the eigenvalues of \( A^{13} \), we need to follow a systematic process:

Step 1: Find the eigenvalues of matrix \( A \)
The eigenvalues of a matrix \( A \) are the solutions to the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \( \lambda \) is an eigenvalue, \( I \) is the identity matrix, and \( \det \) denotes the determinant.
For matrix \( A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \), we calculate: \[ \det(A - \lambda I) = \det\left( \begin{bmatrix} 1-\lambda & 1 \\ 1 & -1-\lambda \end{bmatrix} \right) \] \[ = (1-\lambda)(-1-\lambda) - (1)(1) = \lambda^2 - 1 - 1 = \lambda^2 - 2 \] Now, set the determinant to 0: \[ \lambda^2 - 2 = 0 \] \[ \lambda^2 = 2 \] \[ \lambda = \pm \sqrt{2} \] Thus, the eigenvalues of \( A \) are \( \sqrt{2} \) and \( -\sqrt{2} \).

Step 2: Eigenvalues of \( A^{13} \)
For any matrix \( A \) with eigenvalues \( \lambda_1, \lambda_2, \ldots \), the eigenvalues of \( A^n \) are simply the eigenvalues of \( A \) raised to the power of \( n \). Therefore, the eigenvalues of \( A^{13} \) will be \( (\sqrt{2})^{13} \) and \( (-\sqrt{2})^{13} \).
\[ (\sqrt{2})^{13} = 2^{\frac{13}{2}} = 64\sqrt{2} \] \[ (-\sqrt{2})^{13} = -(2^{\frac{13}{2}}) = -64\sqrt{2} \]
Final Answer:
The eigenvalues of \( A^{13} \) are \( 64\sqrt{2} \) and \( -64\sqrt{2} \).

Thus, the correct answer is (D).