Answer
$\lambda = 0, 1, 3$ with multiplicities of 1, 2, 2, respectively
Work Step by Step
(1) The eigenvalues can be found by taking the determinant of $A-\lambda I$ and set the derivative equal to zero.
$det(A-\lambda I) = (3-\lambda)(1-\lambda)\lambda(1-\lambda)(3-\lambda)=0$
(2) Solve for $\lambda$
(3) $\lambda = 0, 1, 3$ where the eigenvalues of 1 and 3 have multiplicities of two.
