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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.

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May I suggest that you make your title more explicit? –  1015 Apr 7 '13 at 2:06
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2 Answers 2

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

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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 $t^n$. –  1015 Apr 7 '13 at 2:11
@user547866 Is there something you don't understand? –  1015 Apr 7 '13 at 2:17
@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. –  1015 Apr 7 '13 at 2:29
@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. –  1015 Apr 7 '13 at 2:32
@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 $\chi_{AB}=\chi_{BA}$. –  Sami Ben Romdhane Apr 7 '13 at 2:42

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 $\mathbb{R}$. –  Sami Ben Romdhane 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

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