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

Characteristic equation: $(\lambda-3)^2$ $\lambda = 3$

1.) To solve for the characteristic equation, expand and simplify $det(A-\lambda I)$ $(2-\lambda)(4-\lambda)-(1)(-1)$ $=8-2\lambda-4\lambda+\lambda^2+1$ $=\lambda^2-6\lambda+9$ $=(\lambda-3)^2$ 2.) Set the characteristic equation equal to zero to solve for the eigenvalues $(\lambda-3)^2=0$ $\lambda = 3$ Note: The eigenvalue of 3 has a multiplicity of two