# $AB$ and $BA$ have the same characteristic polynomial [closed]

For $A,B$ ∈ $F^{n×n}$, show that $AB$ and $BA$ have the same characteristic polynomial.

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## closed as off-topic by user26857, Leucippus, hardmath, Zachary Selk, G. SassatelliDec 18 '15 at 1:36

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May I suggest that you make your title more explicit? – 1015 Apr 7 '13 at 2:06
It would be a good idea, since you seem to find the site useful, to indicate this by $(1)$ upvoting helpful answers, and $(2)$ accepting helpful answers. You can upvote as many answers as you'd like, but you can accept only one answer per question asked. To accept an answer, just click on the $\large \checkmark$ to the left of the answer you want to accept. It then becomes green. You receive two reputation points whenever you accept an answer. But more importantly, it's a way to show appreciation for the time users take to answer questions that you find helpful. – amWhy Apr 27 '13 at 18:28
– user26857 Dec 17 '15 at 23:52
When asking a question like this, I suggest that you include your ideas or some steps that you tried in order to solve the problem (and where you're stuck) – Michael Burr Dec 18 '15 at 0:37

In the case $F=\mathbb{C}$ or $F=\mathbb{R}$:

First if $A\in\mathrm{GL}_n(F)$ then we have $$\chi_{AB}(x)=\det(xI-AB)=\det(A(xI-BA)A^{-1})=\det(xI-BA)=\chi_{BA}(x)$$ Now by the density of $\mathrm{GL}_n(F)$ in $\mathcal{M}_n(F)$ and the continuity of the $\det$ function we have the desired result.

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You should specify for what fields your argument works. – 1015 Apr 7 '13 at 2:40
@julien You're right I have a bad habit of thinking often for $\mathbb{C}$ or $\mathbb{R}$. – user63181 Apr 7 '13 at 2:48
Well, that's the case most of the time that we are in $\mathbb{C}$ or $\mathbb{R}$ when we do linear algebra. At least at this level. So it is not a real problem, +1. – 1015 Apr 7 '13 at 3:00
Shouldn't this not be a problem? Since both of them are polynomials and proving this on $\mathbb{C}$ proves a polynomial identity, and polynomial identity should hold regardless of rings. – i707107 Apr 7 '13 at 10:29

Edit: There is the argument given by Sami Ben Romdhane which works when we have the density of invertible matrices. I had long known this approach only. But recently I came across this purely algebraic approach which works over any commutative unital ring $R$.

For every $n\times m$ matrix $A$ and every $m\times n$ matrix $B$, we have $$\left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&-A\\0&I}\right)=\left(\matrix{tI&0\\*&tI-BA}\right)$$ and $$\left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&0\\-B&I}\right)=\left(\matrix{tI-AB&*\\0&tI}\right).$$ Taking the determinant of all these yields $$t^m\det(tI-AB)=t^n\det \left(\matrix{I&A\\B&tI}\right)=t^n\det(tI-BA).$$ In the case of square $n\times n$ matrices, this yields, since $n=m$: $$t^n(\det(tI-AB)-\det(tI-BA))=0\quad\Rightarrow\quad \det(tI-AB)-\det(tI-BA)=0$$ where the implication follows from the fact that $\det(tI-AB)-\det(tI-BA)$ is a polynomial in $t$.
Thus $$\chi_{AB}(t)=\det(tI-AB)=\det(tI-BA)=\chi_{BA}(t)\qquad\forall t\in R.$$
@user547866 When $n=m$, there is a factor $t^n$ on both sides: $t^n\det(tI-AB)=t^n\det(tI-BA)$ in the formula I just proved. So I divide by $t^n$. – 1015 Apr 7 '13 at 2:11
@user547866 Then make $m=n$ form the start. It does not change anything. And I don't think it will be easy to find a significantly different way. – 1015 Apr 7 '13 at 2:29
@julien you don't need to treat the case $t=0$ and $t\neq0$ since you deal with polynomials so you can directly conclude that $\chi_{AB}=\chi_{BA}$. – user63181 Apr 7 '13 at 2:42