# Prime X 's question at Yahoo! Answers (Eigenvalues of AB and BA)

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello Prime X,

Suppose $A,B \in\mathbb{F}^{n\times n}$ ($\mathbb{F}$ field) and consider the block matrices $$C=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}\;,\quad D=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}\quad (I\in\mathbb{F}^{n\times n})$$ Then, $$CD=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}=\begin{bmatrix}{-\lambda I+AB}&{-\lambda A}\\{0}&{-\lambda I}\end{bmatrix}\\DC=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}=\begin{bmatrix}{-\lambda I}&{- A}\\{0}&{BA-\lambda I}\end{bmatrix}$$ According to a well-known property $\det(CD)=\det(DC)$ so, $$\det (AB-\lambda I)(-\lambda)^n=(-\lambda)^n\det(BA-\lambda I)$$ Equivalently $$(-\lambda)^n[\det (AB-\lambda I)-\det (BA-\lambda I)]=0$$ But $\mathbb{K}[\lambda]$ is an integral domain so, $$\det (AB-\lambda I)=\det (BA-\lambda I)$$ That is, $AB$ and $BA$ have the same characteristic polynomial (as a consequence the same eigenvalues)

Remark: The above proof also shows that every common eigenvalue has the same multiplicity (with respect to $AB$ and $BA$).