# Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!)

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.

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

## 2 Answers

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 $BA$.

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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
@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 '13 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 '13 at 15:31

It is true that the eigenvalues (counting multiplicity) of $AB$ are the same as those of $BA$.

This is a corollary of Theorem 1.3.22 in the second edition of "Matrix Analysis" by Horn and Johnson, which is Theorem 1.3.20 in the first edition.

Paraphrasing from the cited Theorem: If $A$ is an $m$ by $n$ matrix and $B$ is an $n$ by $m$ matrix with $n \geq m$ then the characteristic polynomial $p_{BA}$ of $BA$ is related to the characteristic polynomial $p_{AB}$ of $AB$ by $$p_{BA}(t) = t^{n-m} p_{AB}(t).$$

In your case, $n = m$, so $p_{BA} = p_{AB}$ and it follows that the eigenvalues (counting multiplicity) of $AB$ and $BA$ are the same.

You can see Horn and Johnson's proof in the Google Books link above. A similar proof was given in this answer from Maisam Hedyelloo.

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