Use the result of Exercise 8 to prove that, if \(A\) and \(B\) are \(n \times n\) matrices over the field \(F\), then \(AB\) and \(BA\) have precisely the same characteristic values in \(F\).
Solution
Let \(c\ne0\). Then \(cI-AB\) is invertible exactly when \(I-c^{-1}AB\) is invertible. Exercise 8, applied to \(c^{-1}A\) and \(B\), says this is equivalent to invertibility of \(I-c^{-1}BA\), hence of \(cI-BA\). Thus \(AB\) and \(BA\) have the same nonzero characteristic values.
For \(c=0\), \(AB\) is invertible iff both \(A\) and \(B\) are invertible. Indeed, invertibility of \(AB\) makes \(B\) injective and \(A\) surjective; for square matrices these imply invertibility. The same criterion holds for \(BA\). Hence \(AB\) is singular iff \(BA\) is singular. Therefore the characteristic-value sets agree completely.