Finding associated eigenvalue and eigenvector Having troubles with this question
Suppose that $\det(A) \not= 0$, and $A$ and $B$ both have eigenvector $v$, but the corresponding eigenvalue is $\lambda_{A}$ for $A$ and $\lambda_{B}$ for $B$. Show that A$^{-1}B$ has the same eigenvector v. Find the eigenvalue associated with this eigenvector for A$^{-1}B$.
Could someone walk me through it?
Thanks
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$$
Bv=\lambda_{B} v\quad\imp\quad
A^{-1}B = \lambda_{B}A^{-1}v;\qquad Av = \lambda_{A}v\quad\imp A^{-1}v = \lambda_{A}^{-1}v
$$
$$
\color{#0000ff}{\large A^{-1}Bv} =\lambda_{B}\pars{\lambda_{A}^{-1}v}
=
\color{#0000ff}{\large{\lambda_{B} \over \lambda_{A}}\,v} 
$$
A: From the very definition of eigenvalues and eigenvectors, you have
$$A\mathbf{v} = \lambda_A\mathbf{v}\tag{1}$$
$$B\mathbf{v} = \lambda_B\mathbf{v}\tag{2}$$
Apply $A^{-1}$ to equation $(1)$ to show that $\mathbf{v}$ is also an eigenvector of $A^{-1}$ but with eigenvalue $\lambda_A^{-1}$. Now substitute equation $(2)$ into $A^{-1}B\mathbf{v}$ and use what we've just learned about $A^{-1}$.
