Are the eigenvalues of the matrix AB equal to the eigenvalues of the matrix BA Are the eigenvalues of the matrix  $
AB
$
 equal to the eigenvalues of the matrix $
BA
$ 
.
Where the matrices A And B of sizes $
{3}\mathrm{\times}{5}
$
and $
{5}\mathrm{\times}{3}
$
Respectively
 .then how can we find the Jordan form of the matrix  $
BA
$
 if we have the matrix:
$
{AB}\mathrm{{=}}\left[{\begin{array}{l}
{{1}\hspace{0.33em}{1}\hspace{0.33em}{0}}\\
{{0}\hspace{0.33em}{1}\hspace{0.33em}{0}}\\
{{0}\hspace{0.33em}{0}\hspace{0.33em}\mathrm{{-}}{1}}
\end{array}}\right]
$
 A: Assuming $A$ and $B$ are square: $BA$ is invertible, so $B$ is invertible. Note that
$$
B^{-1}(BA)B= AB
$$
so $AB$ is similar to $BA$.

Per the clarification: it is well known that $A$ and $B$ will have the same non-zero eigenvalues, as explained in the other answer.  What's more: $AB$ and $BA$ have the same rank, which means that $BA$ (a $5 \times 5$ matrix) has eigenvalue $0$ with algebraic and geometric multiplicity $2$.
We must exclude, however, the possibility that $BA$ is diagonalizable.  Equivalently, we want to show that since $(AB - I)^2$ has a lower rank than $(AB - I)$, $(BA - I)^2$ has a lower rank than $(BA - I)$.

$AB$ has an eigenvector $x$ assoicated with $1$, and a generalized eigenvector $y$ satisfying $ABy = y + x$.  Thus, we see that $BA$ has eigenvector $Bx$, since
$$
(BA)(Bx) = B(AB)x = Bx
$$
moreover, $BA$ has generalized eigenvector $By$ satisfying
$$
(BA)(By) = B(AB)y = B(y + x) = (By) + (Bx)
$$
Thus, $BA$ indeed fails to be diagonalizable.  Thus, we know its Jordan form.
A: Denote $$C=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix},\quad D=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}.$$ Then, $$CD=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}=\begin{bmatrix}{-\lambda I+AB}&{-\lambda A}\\{0}&{-\lambda I}\end{bmatrix},$$ $$DC=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}=\begin{bmatrix}{-\lambda I}&{- A}\\{0}&{BA-\lambda I}\end{bmatrix}.$$ Using that $\det(CD)=\det(DC)$ we get: $$\det (AB-\lambda I)(-\lambda)^n=(-\lambda)^n\det(BA-\lambda I)$$ So $\det(AB-\lambda I)=\det(BA-\lambda I)$ that is,  $AB$ y $BA$ have the same characteristic polynomial as a consequence the same eigenvalues and besides (this is important), with the same algebraic multiplicity.
