Eigenvalues

You should know $\displaystyle \begin{align*} \mathbf{A}\mathbf{x} = \lambda \mathbf{x} \end{align*}$ . Now, if $\displaystyle \begin{align*} \mathbf{A} \end{align*}$ is invertible, then $\displaystyle \begin{align*} \mathbf{A}^{-1} \end{align*}$ exists and $\displaystyle \begin{align*} \mathbf{A}^{-1}\mathbf{A} = \mathbf{I} \end{align*}$ . Anyway...

$\displaystyle \begin{align*} \mathbf{A}\mathbf{x} &= \lambda \mathbf{x} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{x} &= \mathbf{A}^{-1}\lambda \mathbf{x} \\ \mathbf{I}\mathbf{x} &= \lambda \mathbf{A}^{-1}\mathbf{x} \\ \mathbf{x} &= \lambda \mathbf{A}^{-1}\mathbf{x} \\ \frac{1}{\lambda}\mathbf{x} &= \mathbf{A}^{-1}\mathbf{x} \end{align*}$

Since we have the product of two matrices (in this case $\displaystyle \begin{align*} \mathbf{A}^{-1}\mathbf{x} \end{align*}$ ) equal to a scalar multiple of the second matrix (in this case, $\displaystyle \begin{align*} \frac{1}{\lambda}\mathbf{x} \end{align*}$ ), that means that $\displaystyle \begin{align*} \frac{1}{\lambda} \end{align*}$ must be an eigenvalue of $\displaystyle \begin{align*} A^{-1} \end{align*}$ .