Proving that $\lambda$ being an eigenvalue for $A$ implies $\lambda^{-1}$ is an eigenvalue for $A^{-1}$ Let $A$ be an invertible matrix, and let $\lambda$ be an eigenvalue for $A$. We have that $Ax = \lambda x$ for some eigenvector $x$. Note that $A^{-1}Ax = A^{-1}\lambda x$, which gives $x = A^{-1}\lambda x$. Also note that multiplication with scalar is commutative, so this gives $\lambda^{-1}x = A^{-1}x$, which is the desired result.
That last sentence was a realization I made while I was writing the question; so the question has now changed to proof-verification. 
 A: The fact that $x=A^{-1}\lambda x$ does not necessarily mean that $A^{-1}\lambda$ is the identity matrix; what if, for instance, $x$ is the zero-vector?
But, you do have the right idea up until there with your second comment: that $A^{-1}\lambda=\lambda A^{-1}$, so that $\lambda^{-1}x=\lambda^{-1}\lambda A^{-1}x=A^{-1}x$.
A: This is very confuzingly written.
What do you mean when you say $A^{-1}\lambda$ is the identity matrix? $A^{-1}$ is not necesarily diagonal, and $\lambda$ is a scalar, so the matrix is not always the identity...
It is much simpler, since you have $x=A^{-1}\lambda x$ means $x=\lambda A^{-1} x$. Now just multiply this equation with $\lambda^{-1}$ and you are done. On a side note, if I was grading an assignment like this, I would like my students to explain why $\lambda^{-1}$ exists.
A: If $A$ is not defective, you can use the eigendecomposition $A=Q\Lambda Q^{-1}$ and therefore $A^{-1}=Q^{-1}\Lambda^{-1}Q$, where $\Lambda$ contains your eigenvaules and $\Lambda^{-1}$ their reciprocals...
