• Support PF! Buy your school textbooks, materials and every day products Here!

Prove AB and BA have the same characteristic polynomial

  • #1
190
0

Homework Statement


Show that for any square matrices of the same size, A, B, that AB and BA have the same characteristic polynomial.


Homework Equations





The Attempt at a Solution



I understand how to do this if either A or B is invertible, since they would be similar then. I saw a proof that states to take [itex]A = A + \epsilon I[/itex], so that [itex]det(A+\epsilon I) \neq 0[/itex], and then we have,

[tex] det \left((A+\epsilon I) B - \lambda I \right) = det \left((B(A+\epsilon I) - \lambda I\right)[/tex]

Since then we can use the similarity argument. We then take [itex]\epsilon\rightarrow 0[/itex].

I'm wondering if someone could please provide an accessible, but sound explanation of why we are allowed to use this [itex]\epsilon \rightarrow 0 [/itex] argument.

Thanks!
 

Answers and Replies

  • #2
6,054
390
Can you see that the entities on either side of the equation are multivariate polynomials with respect to epsilon and lambda?
 
  • #3
190
0
Hmm, somewhat. Could you please provide greater insight?
 
  • #4
6,054
390
Hmm, somewhat. Could you please provide greater insight?
"Somewhat" is not good. If you really don't see it, take a simple 2x2 case and work it out.

If you do see that, then what do you know of continuity of polynomials? What is [itex]\lim_{\epsilon \rightarrow 0} P(\epsilon, \lambda)[/itex], if [itex]P(\epsilon, \lambda)[/itex] is a polynomial?
 
  • #5
Zondrina
Homework Helper
2,065
136
Hmm are you sure that's the right question? A would need some inverse C such that :

C-1AC = B

Then you could start by saying :

pB(λ) = det(B-λIn)

Then you should use what you know about B and the identity matrix In to massage the expression into pA(λ) [ Hint : Think about inverses and how they work! ]
 

Related Threads on Prove AB and BA have the same characteristic polynomial

Replies
3
Views
15K
Replies
2
Views
3K
Replies
3
Views
2K
Replies
2
Views
3K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
1
Views
16K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
697
Top