### eigenvalues under commutation

$$A$$ and $$B$$ are two square matrices. The eigenvalues of $$AB$$ and $$BA$$ are the same.

Proof: Suppose $$\lambda$$ and $$v$$ are an eigenvalue and the corresponding eigenvector for $$AB$$, so that $$ABv = \lambda v$$. Let $$q = Bv$$. Then $$Aq = \lambda v$$, so $$BAq = \lambda Bv = \lambda q$$. So $$\lambda$$ is also an eigenvalue of $$BA$$, and its associated eigenvector is $$q = Bv$$.