Prove that if y is not zero and is an eigen value of $A$, then $1/y$ is an eigenvalue of the inverse of A I know that to find the eigenvalues of a certain matrix, i need to find the determinant of $A -yI$ ($A$ is the matrix and $I$ is the identity matrix) and then find the zeroes of the quadratic equation 
I also know that i have to prove that the determinant of $A^{-1} - (1/y)I$ is equal to zero, but i'm not sure how to do that exactly
any help will be appreciated
 A: $$\begin{array}{rcl}
Av &=& yv \\
v &=& A^{-1}yv \\
\frac1yv &=& A^{-1}v \\
A^{-1}v &=& \frac1yv
\end{array}$$
A: By applying $A^{-1}$ to both sides of $y v=A v$ and then dividing by $y$ it follows that $A^{-1}y = \frac{1}{y} v.$
A: If $y$ is an eigen value then there exists a vector $v$ such that $Av = yv$ (that's the definition of eigen value).
Then multiply each side by $A^{-1}$:
$$v = A^{-1}Av = A^{-1}yv = yA^{-1}v$$ 
$$\frac{1}{y} v = A^{-1} v$$
Therefore $1/y$ is an eigen value of $A^{-1}$
A: We have$$\det(A-yI)=0$$
Then let's compute$$\begin{align}\det(A^{-1}-\frac1y I)&=\det(A^{-1}A)\det(A^{-1}-\frac1yI)\\&=\det(A^{-1})\det(I-\frac1yA)\\&=y\det(A^{-1})\det(yI-A)\\&=\underbrace{(-1)^ny\det(A^{-1})}_{k}\det(A-yI)\\&=k\det(A-yI)\end{align}$$
where we know $k\ne 0$, and $n$ is the dimension of the space. Then the statement $\text{y is an eigenvalue of }A$ implies the statement $\text{1/y is an eigenvalue of }A^{-1}$ since if $\det(A-yI)=0$, then $\det(A^{-1}-\frac1y I)=0$. 
The key step was multiplying by $\det(A^{-1}A)$, which is just $1$, since it is $\det I$.
A: If
$A \vec v = y \vec v, \; \vec v \ne 0, \tag 1$
that is, $y \ne 0$ is an eigenvalue of $A$ with eigenvector $ \vec v \ne 0$, and $A$ is nonsingular,
then (1) yields
$\vec v = A^{-1}(A \vec v) = A^{-1}(y \vec v) = y(A^{-1}\vec v); \tag 2$
therefore,
$y^{-1} \vec v = y^{-1} y(A^{-1} \vec v) = A^{-1} \vec v, \tag 3$
which shows $y^{-1}$ is an eigenvalue of $A^{-1}$ also with eigenvector $\vec v$.  Also (3) implies
$(A^{-1} - y^{-1} I) \vec v = 0; \tag 4$
(4) leads to the conclusion that
$\det(A^{-1} - y^{-1}) = 0 \tag 5$
as well.
There is another way to see (5), given that
$\det (A - yI) = 0; \tag 6$
we have, from (6),
$\det(y^{-1} A^{-1}) \det(A - yI) = 0; \tag 7$
but
$\det(y^{-1} A^{-1}) \det(A - yI) = \det((y^{-1} A^{-1})(A - yI))= 0; \tag 8$
now since
$(y^{-1} A^{-1})(A - yI) = (y^{-1} A^{-1})A -(y^{-1} A^{-1})y = y^{-1} I - A^{-1}, \tag 9$
we see that (7) leads to
$\det(A^{-1} - y^{-1}I) = 0, \tag{10}$
which shows that $y^{-1}$ satisfies the characteristic polynomial of $A^{-1}$.
These considerations need not be restricted to matrices of size $2$; they hold for matrices of any size and indeed over any field of scalars.
