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

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For n-square matrices A and B, it is proven that the matrices AB and BA share the same eigenvalues. This is demonstrated using block matrices C and D, leading to the conclusion that the determinants of the characteristic polynomials of AB and BA are equal. The proof relies on the property that the determinant of the product of matrices is invariant under the order of multiplication. Consequently, both matrices have the same characteristic polynomial, which implies they possess identical eigenvalues. Additionally, the proof indicates that any common eigenvalue retains the same multiplicity for both matrices.
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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 $BA$).
 
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