Chapter 5 - Eigenvalues and Eigenvectors - 5.2 Exercises - Page 281: 2

Characteristic equation: $\lambda^2-10\lambda+16$ $\lambda=2,8$

1.) To find the characteristic polynomial, find $det(A-\lambda I)$ $=(5-\lambda)(5-\lambda)-9$ $=\lambda^2-10\lambda+16$ 2.) Solve for the eigenvalues by equating the characteristic equation to zero and solving for $\lambda$ $\lambda^2-10\lambda+16=0$ $(\lambda-2)(\lambda-8) = 0$ $\lambda=2,8$