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Can someone please help me with this proof.

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

user26857
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user547866
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    May I suggest that you make your title more explicit? – Julien Apr 07 '13 at 02:06
  • 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
  • Related: http://math.stackexchange.com/questions/311342/do-ab-and-ba-have-same-minimal-and-characteristic-polynomials/311362#311362 – 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 00:37

2 Answers2

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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.

  • You should specify for what fields your argument works. – Julien Apr 07 '13 at 02:40
  • @julien You're right I have a bad habit of thinking often for $\mathbb{C}$ or $\mathbb{R}$. –  Apr 07 '13 at 02:48
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    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. – Julien Apr 07 '13 at 03:00
  • 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. – Sungjin Kim Apr 07 '13 at 10:29
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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 $R$.

I'll start with a slightly more general result for rectangular matrices.

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 $$ \chi_{AB}(t)=\det(tI-AB)=\det(tI-BA)=\chi_{BA}(t)\qquad\forall t\in R. $$

Julien
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  • @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 $t^n$. – Julien Apr 07 '13 at 02:11
  • @user547866 Is there something you don't understand? – Julien Apr 07 '13 at 02:17
  • @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. – Julien Apr 07 '13 at 02:29
  • @user547866 If you go this kind of way, you will only show that the eigenvalues are the same. You will not deduce anything on the their multiplicities. And this won't tell you anything about the non linear factors eather. For that, you would have to go in the algebraic closure of the field. But anyway: you will miss the multiplicities. – Julien Apr 07 '13 at 02:32
  • @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 $\chi_{AB}=\chi_{BA}$. –  Apr 07 '13 at 02:42
  • @SamiBenRomdhane Right. I made my life more complicated. Thanks. – Julien Apr 07 '13 at 02:54
  • @user547866 That's a formula. It is a matrix trick. I've learned it. I did not invent it. If you introduce these block matrices, you compute, and you get the result. – Julien Apr 07 '13 at 02:58