Let \(A\) and \(B\) be \(n \times n\) matrices over the field \(F\). According to Exercise 9 of Section 6.1, the matrices \(AB\) and \(BA\) have the same characteristic values. Do they have the same characteristic polynomial? Do they have the same minimal polynomial?
Solution
They always have the same characteristic polynomial. Over the polynomial ring \(F[x]\), regard \(xI-A B\) and \(xI-B A\) as matrices over the field \(F(x)\). Since \(x\ne0\) in \(F(x)\), \[ \det(xI-AB)=x^n\det(I-x^{-1}AB). \] For square matrices \(C,D\) over any field, the identity \(\det(I-CD)=\det(I-DC)\) follows directly from the block matrix \[ M=\begin{bmatrix}I&C\\D&I\end{bmatrix}. \] Indeed, the two block matrices \[ L_1=\begin{bmatrix}I&-C\\0&I\end{bmatrix}, \qquad L_2=\begin{bmatrix}I&0\\-D&I\end{bmatrix} \] have determinant \(1\), while \[ L_1M=\begin{bmatrix}I-CD&0\\D&I\end{bmatrix}, \qquad L_2M=\begin{bmatrix}I&C\\0&I-DC\end{bmatrix}. \] Taking determinants gives \(\det(I-CD)=\det M=\det(I-DC)\). Applying this with \(C=x^{-1}A\) and \(D=B\) gives \[ \det(xI-AB)=x^n\det(I-x^{-1}BA)=\det(xI-BA). \] Both sides lie in \(F[x]\), so the characteristic polynomials agree.
The minimal polynomials need not agree. Take \[ A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\qquad B=\begin{bmatrix}0&1\\0&0\end{bmatrix}. \] Then \[ AB=\begin{bmatrix}0&1\\0&0\end{bmatrix},\qquad BA=0. \] Thus \((AB)^2=0\) but \(AB\ne0\), so \(m_{AB}(x)=x^2\), whereas \(m_{BA}(x)=x\). Hence equal characteristic polynomials do not force equal minimal polynomials.