Let's solve the problem step by step: --- ### **Step 1: Find the Eigenvalues** Given matrix: \[ A = \begin{bmatrix} 5 & 2 \\ -5 & -1 \end{bmatrix} \] The characteristic equation is: \[ \det(A - \lambda I) = 0 \] So, \[ \begin{vmatrix} 5 - \lambda & 2 \\ -5 & -1 - \lambda \end{vmatrix} = 0 \] \[ (5 - \lambda)(-1 - \lambda) - (2)(-5) = 0 \] \[ (5 - \lambda)(-1 - \lambda) + 10 = 0 \] \[ (5 - \lambda)(-1 - \lambda) = -5 - 5\lambda + \lambda + \lambda^2 = \lambda^2 - 4\lambda - 5 \] So: \[ \lambda^2 - 4\lambda - 5 + 10 = 0 \] \[ \lambda^2 - 4\lambda + 5 = 0 \] Solve quadratic: \[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i \] **End of Step 1 Explanation:** The eigenvalues are \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \). --- ### **Step 2: Find the Corresponding Eigenvectors** Let's find the eigenvector for \( \lambda_1 = 2 + i \): Solve \((A - \lambda_1 I)\vec{v} = 0\): \[ A - \lambda_1 I = \begin{bmatrix} 5 - (2+i) & 2 \\ -5 & -1 - (2+i) \end{bmatrix} = \begin{bmatrix} 3-i & 2 \\ -5 & -3-i \end{bmatrix} \] Let \(\vec{v}_1 = \begin{bmatrix} x \\ y \end{bmatrix}\), so: \[ (3-i)x + 2y = 0 \implies y = \frac{-(3-i)}{2}x \] Let \(x = 1\), then \(y = \frac{-(3-i)}{2}\). So, \[ \vec{v}_1 = \begin{bmatrix} 1 \\ \frac{-(3-i)}{2} \end{bmatrix} \] Similarly, for \( \lambda_2 = 2 - i \): \[ A - \lambda_2 I = \begin{bmatrix} 3+i & 2 \\ -5 & -3+i \end{bmatrix} \] \[ (3+i)x + 2y = 0 \implies y = \frac{-(3+i)}{2}x \] Let \(x = 1\), then \(y = \frac{-(3+i)}{2}\). So, \[ \vec{v}_2 = \begin{bmatrix} 1 \\ \frac{-(3+i)}{2} \end{bmatrix} \] **End of Step 2 Explanation:** We found the eigenvectors corresponding to each eigenvalue. --- ### **Final Answers:** \[ \boxed{ \lambda_1 = 2 + i, \quad \vec{v}_1 = \begin{bmatrix} 1 \\ \frac{-(3-i)}{2} \end{bmatrix} } \] \[ \boxed{ \lambda_2 = 2 - i, \quad \vec{v}_2 = \begin{bmatrix} 1 \\ \frac{-(3+i)}{2} \end{bmatrix} } \]