Let A,B be two square matrices of order n. Do AB and BA have same minimal and characteristic polynomials?
I have a proof only if A or B is invertible. Is it true for all cases?
Answer
If A is invertible then A−1(AB)A=BA, so AB and BA are similar, which implies (but is stronger than) AB and BA have the same minimal polynomial and the same characteristic polynomial.
The same goes if B is invertible.
In general, from the above observation, it is not too difficult to show that AB, and BA have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
If the matrices are in Mn(C), you use the fact that GLn(C) is dense in Mn(C) and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than C).
In general AB and BA do not have the same minimal polynomial. I'll let you search a bit for a counter example.
No comments:
Post a Comment