Why is $\frac{1}{\lambda}$ the inverse of $\lambda$? This is the question I'm working on:

Suppose that $\lambda$ is an eigenvalue for $\phi \in \mbox{End}(v)$. Show that $\lambda ^{-1}$ is an eigenvalue for $\phi ^{-1}$.

I know to solve I simply do:
$\phi v = \lambda v$ then apply $\phi ^{-1}$ to both sides and get
$(\phi ^{-1}) \phi v = \lambda v (\phi ^{-1})$ and since $\phi\ \phi ^{-1}$ is the identity it comes to 
$v = \phi ^{-1} \lambda v$ divide both sides by $\lambda$ and get
$$\frac{1}{\lambda} v = \phi ^{-1} v$$
And so $\frac{1}{\lambda}$ is the eigenvalue of $\phi ^{-1}$.
But my question is why is $\frac{1}{\lambda}$ the inverse of $\lambda$? Is this just a known fact or is there a way to prove this statement?
 A: First note that your statement
$$
(\phi ^{-1}) \phi v = \lambda v (\phi ^{-1})
$$
must be write as
$$
\phi ^{-1} \phi (v) = \lambda \phi ^{-1}(v)
$$
this is not a pedantic observation, because a good notation is an important help to well understand the meaning. And this means that $\lambda$ is an element of the field over which the vector space where $\phi$ operate is defined. So since in a field any nonzero element has an unique inverse, there is an inverse of $\lambda$, that usually is write as $\lambda^{-1}$ or $\frac{1}{\lambda}$, and has the property (by definition) that $\frac{1}{\lambda} \lambda =1$. So, multiplying the two side of the equation by this inverse you have:
$$
\frac{1}{\lambda} v=\phi^{-1}(v)
$$
A: However you define the operations between numbers, $1/\lambda$ is the number that, multiplied by $\lambda$, gives you one; that's what an inverse is. 
A: In a field $K$ you define $x/y$ ($y\neq 0$) by $xy^{-1}$, where $y^{-1}$ is the inverse of $y$ in the abelian group $(K\setminus\{0\},\cdot)$. Since $1$ is the neutral element of that group, you have $1/y = 1\cdot y^{-1} = y^{-1}$, which is the inverse of $y$.
A: Let $A\mathbf{X}=\lambda\mathbf{X}$
Pre-multiply by $A^{-1}$
$A^{-1}A\mathbf{X}=A^{-1}\lambda\mathbf{X}$
$\mathbf{X}=\lambda A^{-1}\mathbf{X}$
$\frac{1}{\lambda}=A^{-1}\mathbf{X}$
Hence e/value of $A^{-1}$ is $\frac{1}{\lambda}$, which answers the question
