Solution
= Solution
From $A\mathbf v=\lambda B\mathbf v$ we have $(A-\lambda B)\mathbf v=0$ with $\mathbf v\ne0$. Thus $A-\lambda B$ is singular and \b[$\det(A-\lambda B)=0$].
= Solution
From $A\mathbf v=\lambda B\mathbf v$ we have $(A-\lambda B)\mathbf v=0$ with $\mathbf v\ne0$. Thus $A-\lambda B$ is singular and \b[$\det(A-\lambda B)=0$].