Let $\lambda$ be an eigenvalue of $A$. Prove that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$. 
Let $\lambda$ be an eigenvalue of $A$. Prove that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

My approach:
Suppose $\lambda$ is an eigenvalue of $A$. Then $Ax=\lambda x$ for some $x\neq 0$. Since $A$ is invertible, $Ax=\lambda x \implies A^{-1}Ax=A^{-1}\lambda x$. So:
$$Ix=A^{-1}\lambda x \iff Ix-A^{-1}\lambda x=0\iff x(I-A^{-1}\lambda)=0 $$
Since $x \neq 0$ and $A$ is invertible, $\lambda^{-1}=A^{-1}$ (I'm unsure of this equality). I think this somewhat shows it, this might seem elementary but aren't $\lambda, \lambda^{-1}$ scalars? So basically, if this proof is true $\lambda\cdot\lambda^{-1}=I$ also?
Any other way to show this? Thanks
 A: Your proof is fine, but you seem to be making things a bit overly complicated.
Proof: Let $x \neq 0$ be such that $Ax = \lambda x$.  Since $A$ is invertible, it has a trivial kernel, so $\lambda \neq 0$. We then have
$$
Ax = \lambda x \implies\\
A^{-1}Ax = A^{-1}(\lambda x) \implies\\
x = \lambda A^{-1}(x) \implies\\
\frac 1{\lambda} x = A^{-1}x
$$
By the above equation, $x$ is an eigenvector of $A^{-1}$ associated with the eigenvalue $1/\lambda$. The conclusion holds.
A: (As Robert Israel mentioned in comments, you must assume $A$ is invertible).
There is a more direct approach you can take.  As you said
$$Ax=\lambda x \implies A^{-1}Ax=A^{-1}\lambda x$$
and then
$$A^{-1}Ax=A^{-1}\lambda x \implies Ix = A^{-1}(\lambda x) = \lambda A^{-1}x$$
Now dividing both sides by $\lambda$ yields
$$\lambda^{-1}x = A^{-1}x$$
i.e. $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.
A: Before I continue, it is important to note that $\lambda$ is a scalar and not a matrix. So given that $A$ is invertible, $Ax=\lambda x$, $A$ is invertible, and $\lambda\neq 0$, we have
$$Ax=\lambda x\implies A^{-1}Ax=A^{-1}\lambda x\implies x=\lambda A^{-1}x\implies \frac1\lambda x=A^{-1}x.$$
A: *

*Your worry about equating scalars to matrices is justified. Don't do it! 
If we had $I-A^{-1}\lambda=0$, then multiplying by $\lambda^{-1}$ (also called $\frac1\lambda$) we would get $\lambda^{-1}I-A^{-1}=0$, i.e. $\lambda^{-1}I=A^{-1}$, which are now two matrices.

*Matrix multiplication keeps less properties of number multiplication, e.g. the implication 
'$AB=0\implies A=0$ or $B=0$' $\,$ is lost, so we won't get the conclusion $I-A^{-1}\lambda=0$.

*All we need is
$$A^{-1}x=\lambda^{-1}x$$
for this particular $x$.

A: From $Ix=A^{-1}\lambda x$ you have $A^{-1}x=\lambda^{-1}x$, which shows what you want. You are correct that $\lambda^{-1}$ is a scalar, so cannot equal $A^{-1}$. Their products with the vector $x$ are equal, but that does not say they are equal.
A: In general we have the equation

$$
\sum_{k=0}^n a_k \mathbf{A}^k = 0. \tag 1
$$

The characteristic polynomial for the matrix $\mathbf{A}$ is then given by

$$
\sum_{k=0}^n a_k x^k = 0. \tag 2
$$

The zeros of this equation are the eigenvalues.
If no eigenvalue is $0$, then $\mathbf{A}^{-1}$ is defined.
So we can write

$$
\mathbf{A}^{-n} \sum_{k=0}^n a_k \mathbf{A}^{k}
= \sum_{k=0}^n a_k \mathbf{A}^{k-n}
= \sum_{k=0}^n a_{n-k} \Big( \mathbf{A}^{-1} \Big)^{k} = 0. \tag 3
$$

The characteristic polynomial for the matrix $\mathbf{A}^{-1}$ is then given by

$$
\sum_{k=0}^n a_{n-k} \big( x' \big) ^k = 0. \tag 4
$$

Then

$$
\Bigg( \frac{1}{x'} \Bigg)^{-n} \sum_{k=0}^n a_{n-k} \big( x' \big) ^k
= \sum_{k=0}^n a_{n-k} \Bigg( \frac{1}{x'} \Bigg)^{n-k}
= \sum_{k=0}^n a_k \Bigg( \frac{1}{x'} \Bigg)^k = 0. \tag 5
$$

Compare (2) and (5) and we see that the roots satisfy

$$
x' = \frac{1}{x}. \tag 6
$$

Conclusion:

So IF $\lambda$ is an eigenvalue of the invertible matrix $\mathbf{A}$
THEN $\lambda^{-1}$ an eigenvalue of the matrix $\mathbf{A}^{-1}$.

A: You're making it too hard. Say $x\ne0$ and $Ax=\lambda x$. 
Then $x=A^{-1}Ax=\lambda A^{-1}x$, so $\lambda^{-1}x=A^{-1}x$.
Details: Of course we need to assume here that $A$ is invertible. And then we need to show that it follows that $\lambda\ne0$, which is easy: If $0$ is an eigenvalue of $A$ then $Ay=0$ for some $y\ne0$, showing that $A$ is not invertible.
