Are the eigenvalues of the matrix AB equal to the eigenvalues of the matrix BA

Question

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] $

Answer 1

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 $BA$.

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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 $(BA - I)$.

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$AB$ 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.

Answer 2

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.