47
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you tried to find a counter example using basic 2x2 matrices? $\endgroup$ – Marra Mar 27 '12 at 1:16
82
$\begingroup$

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 $BA$.

$\endgroup$
  • 1
    $\begingroup$ Bob, I am a little confused on your proof. How did you know to use the trick $Bv \ne 0$ to prove this? $\endgroup$ – diimension Nov 24 '12 at 1:54
  • 4
    $\begingroup$ @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. $\endgroup$ – Phira Dec 21 '12 at 17:27
  • 1
    $\begingroup$ @Phira thank you very much! $\endgroup$ – diimension Dec 26 '12 at 8:21
  • 2
    $\begingroup$ 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 $\lambda=0$? $\endgroup$ – crf Apr 22 '13 at 12:32
  • 2
    $\begingroup$ @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. $\endgroup$ – Bob Pego May 7 '13 at 15:31
15
$\begingroup$

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.

$\endgroup$
13
$\begingroup$

Here is an alternative proof for this result, following Exercises 6.2.8-9 of Hoffman & Kunze's Linear Algebra (p. 190):


Lemma: Let $A,B\in M_n(\mathbb{F})$, where $\mathbb{F}$ is an arbitrary field. If $I-AB$ is invertible, then so is $I-BA$, and

$$(I-BA)^{-1}=I+B(I-AB)^{-1}A.$$

Proof of Lemma: Since $I-AB$ is invertible,

\begin{align} &I=(I-AB)(I-AB)^{-1}=(I-AB)^{-1}-AB(I-AB)^{-1}\\ &\implies (I-AB)^{-1} = I+ AB(I-AB)^{-1}. \end{align}

Then we have

\begin{align} I+B(I-AB)^{-1}A&= I+B[I+ AB(I-AB)^{-1}]A= I+BA+BAB(I-AB)^{-1}A\\ \implies I&=I+B(I-AB)^{-1}A-BA-BAB(I-AB)^{-1}A\\ &=I[I+B(I-AB)^{-1}A]-BA[I+B(I-AB)^{-1}A]\\ &=(I-BA)[I+B(I-AB)^{-1}A].\checkmark. \end{align}


Proposition: $\forall A,B\in M_n(\mathbb{F}):$ $AB$ and $BA$ have the same eigenvalues.

Proof: Let $\alpha\in\mathbb{F}$ be an eigenvalue of $AB$. If $\alpha=0$, then $0=\det(0I-AB)=\det(-A)\det(B)=\det(B)\det(-A)=\det(0I-BA)$ and so $0$ is an eigenvalue of $BA$ also.

Otherwise $\alpha\neq0$. Suppose $\alpha$ is not an eigenvalue of $BA$. Then $0\neq\det(\alpha I-BA)=\alpha^n\det(I-(\frac{1}{\alpha}B)A)$. Then $0\neq\det(I-(\frac{1}{\alpha}B)A),$ so that $I-(\frac{1}{\alpha}B)A$ is invertible. By the lemma above we know that $I-A(\frac{1}{\alpha}B)$ is invertible as well, meaning $0\neq\det(I-A(\frac{1}{\alpha}B))=\det(I-\frac{1}{\alpha}AB) \implies 0\neq\det(\alpha I-AB)$. But we assumed $\alpha$ to be an eigenvalue for $AB$, $\unicode{x21af}$.

$\endgroup$
  • $\begingroup$ Do we need the condition of invertibility of $A$ or $B$ here? $\endgroup$ – StubbornAtom Dec 29 '16 at 10:46
  • $\begingroup$ @StubbornAtom No, the arguments I presented make no use of that condition ($M_n(\Bbb{F})$ is the set of all $n\times n$ matrices with entries from $\Bbb{F}$). $\endgroup$ – Alp Uzman Dec 30 '16 at 17:11
  • 2
    $\begingroup$ Amazing answer. Thank you.. $\endgroup$ – Hirak Jun 4 '17 at 7:38
  • $\begingroup$ How u get $det(αI- BA) = α^n$ ? $\endgroup$ – Akash Patalwanshi Oct 11 '17 at 6:51
  • $\begingroup$ The equation doesn't end there, and determinant is multilinear on columns. $\endgroup$ – Alp Uzman Oct 12 '17 at 0:27
4
$\begingroup$

Notice that $\lambda$ being an Eigen value of $AB$ implies that $\det(AB-\lambda I)=0$ which implies that $$\det(A^{-1})\det(AB-\lambda I)\det(B^{-1})=0=\det(A^{-1}(AB-\lambda I)B^{-1})=\det((B-\lambda A^{-1})B^{-1}) $$ $$=\det(I-\lambda A^{-1}B^{-1}) = 0$$ This further implies that $$\det(BA)\det(I-\lambda A^{-1}B^{-1})=\det(BA(I-\lambda A^{-1}B^{-1}))=\det(BA-\lambda I)=0$$ i.e. $\lambda$ is an Eigen value of $BA$. This proof holds only for invertible matrices $A$ and $B$ though. For singular matrices you can show that $0$ is a common Eigen value but I can't think of a way to show that the rest of the Eigen values are equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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