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