Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (a)/Hint 1

Recall that an eigenvector $\mathbf {x}$ of an eigenvalue $\lambda$ satisfies the equation ${\textstyle A\mathbf {x} =\lambda \mathbf {x} }$, or equivalently, ${\textstyle (A-\lambda I)\mathbf {x} =\mathbf {0} }$.