Question

Consider the matrix: \[ \begin{bmatrix} 1 & k \\ 2 & 1 \end{bmatrix}, \] where \(k\) is a positive real number. Which of the following vectors is/are eigenvector(s) of this matrix?

• A. \[ \begin{bmatrix} 1 \\ -\sqrt{2}/k \end{bmatrix} \]
• B. \[ \begin{bmatrix} 1 \\ \sqrt{2}/k \end{bmatrix} \]
• C. \[ \begin{bmatrix} \sqrt{2k} \\ 1 \end{bmatrix} \]
• D. \[ \begin{bmatrix} \sqrt{2k} \\ -1 \end{bmatrix} \]

Hint:

To verify eigenvectors, substitute them into the equation \(A \vec{v} = \lambda \vec{v}\) and check for consistency.

Answer 1

The Correct Option is A

Step 1: Calculate eigenvalues.
For the given matrix: \[ A = \begin{bmatrix} 1 & k \\ 2 & 1 \end{bmatrix} \] The characteristic equation is determined as: \[ \det \begin{bmatrix} 1 - \lambda & k \\ 2 & 1 - \lambda \end{bmatrix} = 0 \] Expanding the determinant: \[ (1 - \lambda)^2 - 2k = 0 \] Simplify to: \[ \lambda^2 - 2\lambda - 2k + 1 = 0 \] Solve for \(\lambda\): \[ \lambda = 1 \pm \sqrt{2k} \]

Step 2: Derive eigenvectors.
For eigenvalue \(\lambda_1 = 1 + \sqrt{2k}\): \[ \begin{bmatrix} 1 - (1 + \sqrt{2k}) & k \\ 2 & 1 - (1 + \sqrt{2k}) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0 \] Simplify to find the ratio \(\frac{y}{x}\): \[ y = \frac{\sqrt{2}}{k} x \] Thus, an eigenvector corresponding to \(\lambda_1\) is: \[ \begin{bmatrix} 1 \\ \frac{\sqrt{2}}{k} \end{bmatrix} \]

For eigenvalue \(\lambda_2 = 1 - \sqrt{2k}\): \[ y = -\frac{\sqrt{2}}{k} x \] Thus, an eigenvector corresponding to \(\lambda_2\) is: \[ \begin{bmatrix} 1 \\ -\frac{\sqrt{2}}{k} \end{bmatrix} \] Therefore, the correct eigenvectors are option (1) and (2). Final Answer: \[ \boxed{(1), (2)} \]