Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite
3

First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?

If it's not true, then under what conditions is it true or not true?

If it is true, can anyone point me to a citation? I couldn't find it in a quick perusal of Horn & Johnson. I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.

A trivial proof would be OK, but a citation is better.

share|improve this question
Have you tried to find a counter example using basic 2x2 matrices? – Gustavo Marra Mar 27 '12 at 1:16

1 Answer

up vote 13 down vote accepted

If $v$ is an eigenvector of $AB$ for some nonzero $\lambda$, then $Bv\ne0$ and $$\lambda Bv=B(ABv)=(BA)Bv,$$ so $Bv$ is an eigenvector for $BA$ with the same eigenvalue. If $0$ is an eigenvalue of $AB$ then $0=\det(AB)=\det(A)\det(B)=\det(BA)$ so $0$ is also an eigenvalue of $BA$.

More generally, Jacobson's lemma in operator theory states that for any two bounded operators $A$ and $B$ acting on a Hilbert space $H$ (or more generally, for any two elements of a Banach algebra), the non-zero points of the spectrum of $AB$ coincide with those of the spectrum of $BA$.

share|improve this answer
Bob, I am a little confused on your proof. How did you know to use the trick $Bv \ne 0$ to prove this? – diimension Nov 24 '12 at 1:54
1  
@diimension You just take an eigenvector of $AB$ and do the only calculation you can. Then it turns out that $Bv$ fulfills the eigenvector equation for $BA$, so you hope that it is not 0 and check this in the end. There's no need to know it at the start of the calculation. – Phira Dec 21 '12 at 17:27
@Phira thank you very much! – diimension Dec 26 '12 at 8:21
I know this is an ancient thread but hopefully you're still lurking out there somewhere. How come you need to hope that $\lambda\ne 0$? Doesn't the argument still work just fine in the case where $\lambda=0$? – crf Apr 22 at 12:32
@crf No -- the trouble is that when $Bv=0$, maybe $v$ is not in the range of $A$. The argument given for $\lambda\ne0$ works in Hilbert space, for example, but for $\lambda=0$ the result is not generally true there: On the infinite sequence space $\ell^2$, let $A$ be the right shift ($A(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$, and $B$ the left shift. Then $AB(1,0,\ldots)=0$, but $BA$ is the identity. – Bob Pego May 7 at 15:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.