Question

Let \( M = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \) and \( K = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \) satisfy the eigenvalue problem given by:
\[ (M - \alpha K)\phi = 0. \] The lowest eigenvalue \( \alpha \) is ________ (rounded off to two decimal places).

Hint:

When solving eigenvalue problems, remember to set the determinant of \( (M - \alpha K) \) to zero. The solutions to the resulting equation give the eigenvalues.

Answer 1

We are given the matrix equation \( (M - \alpha K)\phi = 0 \), which is the standard form of an eigenvalue problem. This implies that for nontrivial solutions \( \phi \), the matrix \( (M - \alpha K) \) must be singular, i.e., its determinant must be zero. Thus, we need to solve: \[ \det(M - \alpha K) = 0. \] First, calculate \( M - \alpha K \): \[ M - \alpha K = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \alpha \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 2\alpha & \alpha \\ \alpha & 2 - \alpha \end{bmatrix}. \] Now, compute the determinant of \( M - \alpha K \): \[ \det(M - \alpha K) = \det \begin{bmatrix} 1 - 2\alpha & \alpha \\ \alpha & 2 - \alpha \end{bmatrix} = (1 - 2\alpha)(2 - \alpha) - \alpha^2. \] Expanding the terms: \[ = (1 - 2\alpha)(2 - \alpha) - \alpha^2 = 2 - \alpha - 4\alpha + 2\alpha^2 - \alpha^2 = 2 - 5\alpha + \alpha^2. \] To find the eigenvalues, solve the quadratic equation: \[ \alpha^2 - 5\alpha + 2 = 0. \] Using the quadratic formula: \[ \alpha = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}. \] Thus, the eigenvalues are: \[ \alpha = \frac{5 + \sqrt{17}}{2} \quad \text{or} \quad \alpha = \frac{5 - \sqrt{17}}{2}. \] The lowest eigenvalue is: \[ \alpha = \frac{5 - \sqrt{17}}{2} \approx 0.44. \]