The Eigenvalues of AB are the Eigenvalues of BA

We prove that for two square matrices A and B of the same dimension, the eigenvalues of AB are the same as the eigenvalues of BA. We use a variant of a Neumann series argument to explicitly construct the inverse of I-AB when it exists and relate it to the inverse of I-BA.
#mikethemathematician, #mikedabkowski, #profdabkowski

[@punditgi]

Magic Mike! 🎉😊

[@mikethemathematician]

@punditgi Thanks so much Ezra!