Prove: $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ $\implies$ $v$ is an eigenvector of $A^{-1}$ with an eigenvalue of $\frac{1}{\lambda}$

Question

I know that $$det(AA^{-1})=det(I)=1$$ And because $$det(AA^{-1})=det(A)det(A^-1)$$ So: $$det(A^{-1})=\frac{1}{det(A)}$$ But not sure how to go from here and if that's even useful, because $A-\lambda I$ is singular and there is no way to express the sum of 2 matrices (Expressing the determinant of a sum of two matrices?).

(Clues would be more appreciated than solutions)

Answer 1

$Av=\lambda v$ and if $A^{-1}$ exists.then we can operate $A^{-1}$ in both sides to get $$v=A^{-1}\lambda v$$ $$\frac{v}{\lambda}=A^{-1}v$$

Answer 2

Hint: Start from $Av = \lambda v$ and multiply by $A^{-1}$ from the left.

Answer 3

$$Av=\lambda v$$ Multiply $A^{-1}$ from the left on both sides: $$A^{-1}Av=Iv=v=A^{-1}\lambda v$$ Multiply both sides by $\frac{1}{\lambda}$: $$\frac{1}{\lambda}v=\frac{1}{\lambda}(A^{-1}\lambda v)=\frac{1}{\lambda}(\lambda A^{-1}v) = A^{-1}v$$ Proving that $\frac{1}{\lambda}$ is an eigenvalue of $v$ for $A^{-1}$