Eigenvalues of a Symmetric 2x2 Matrix

Solve $\det (A-\lambda I)=0$ for this symmetric matrix.

The characteristic polynomial is $(3-\lambda {)}^2-1=0$.

1. 1. Set up the characteristic equation

   The eigenvalues satisfy $\det (A-\lambda I)=0$.

   $$(3-\lambda {)}^2-1=0$$

2. 2. Solve for lambda

   Expanding gives ${\lambda }^2-6\lambda +8=0$, which factors as $(\lambda -4)(\lambda -2)=0$, so the eigenvalues are $4$ and $2$.

Answer: 4 and 2