Register to reply

What are the simplest way to show that AB and BA have the same characteristic polyno?

by td21
Tags: characteristic, polyno, simplest
Share this thread:
Jun11-11, 03:59 AM
td21's Avatar
P: 104
i find this but i do not understand;its too complex.

any simpler way/idea?Thanks!
Phys.Org News Partner Science news on
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
Jun11-11, 05:58 AM
P: 313
If you're happy that Det(AB)=Det(BA), then with I the identity matrix,
the characteristic polynomial is

p(x) = Det(AB-xI) = Det(IAB-xI2) = Det(B-1BAB-xB-1BI)
= Det(BABB-1-xBIB-1) = Det(BA-xI)
Jun11-11, 07:16 AM
P: 8
If A is invertible, than it is evident that AB and BA are similar matrices, therefore they have the same characteristic polynomial. Otherwise, we notice that the equation in lambda, det(A-lambda I)=0 has finitely many solutions. We can take epsilon such that, for all lambda 0<|lambda|<epsilon, A-lambda I is invertible. Therefore, (A-lambda I)B and B(A-lambda I) must have the same characteristic equation. Also, det(xI-(A-lambda I)B)=det(xI-B(A-lambda I)). For fixed x, each side of this equation is a polynomial in lambda, hence it is continuous. We can take lambda->0 and we are done.

Register to reply

Related Discussions
How does one show that the fat cantor characteristic function is nonriemann int'able? Calculus 5
The simplest of questions. Cosmology 6
Simplest fat Biology 4
Simplest question ever Engineering Systems & Design 11
Simplest way to pay debts Fun, Photos & Games 0