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Suppose $A B$ are all $n\times n$ matrix. Prove:$A B$ and $B A$ has the same characteristic polynomial.

  • If A is invertible

\begin{align}P(A B)=|\lambda E-A B|=\left|\lambda A\cdot A^{-1}-A B\right|=\left|A\left(\lambda A^{-1}-B\right)\right|=\left|A\left\|\text{$\lambda $A}^{-1}-B\right.\right|\end{align}

\begin{align}P(B A)=|\lambda E-B A|=\left|\lambda A^{-1}\cdot A-B A\right|=\left|\left(\lambda A^{-1}-B\right)A\right|=\left|\left.\text{$\lambda $A}^{-1}-B\right\|A\right|\end{align}

Is it true?

  • If A is not invertible

How to prove?

Edit:

I found a proof by Block Matrix Multiplication, it's more straightforward, is it right?

\begin{align}\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right).\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E & B \\ 0 & \lambda E-A B \\ \end{array} \right)\end{align}

\begin{align}\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right).\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E-B A & B \\ 0 & \lambda E \\ \end{array} \right)\end{align}

Another proof is by Sylvester determinant theorem which uses LU block decomposition.

integer
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3 Answers3

19

Fill two matrices $A$ and $B$ with $2n^2$ distinct indeterminates. Observe $A$ and $B$ are invertible in the field $F=K(a_{ij},b_{ij})$ formed by adjoining these formal variables. Hence $\chi_{AB}(T)=\chi_{BA}(T)$ holds in the field $F[T]$ and so in the subring $K[a_{ij},b_{ij}][T]$, after which we can simply apply the evaluation map so that $\chi_{AB}(T)=\chi_{BA}(T)$ holds for any two matrices $A$ and $B$ with entries taken in $K$.

The beauty of this "universal" argument is that it bypasses the analytic concept of density, instead working purely algebraically and applying to any desired field $K$.

anon
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  • However, a dense set in the Zariski topology is implicitly used to show that $A$ and $B$ are invertible matrices over $F$ (their determinants being nonzero). – Marc van Leeuwen Oct 06 '13 at 06:26
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    This is like using an atomic bomb to kill a fly. – Raskolnikov Oct 06 '13 at 06:27
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    @Raskolnikov The tools used would be considered flies in the broader context of modern algebra. – anon Oct 06 '13 at 06:28
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    @MarcvanLeeuwen Would you say the fact that $x$ is nonzero in $K(x)$, where $x$ is an indeterminate, uses Zariski topology implicitly? – anon Oct 06 '13 at 06:29
  • @Raskolnikov This reminds me of the legal proverb "Die Polizei soll nicht mit Kanonen auf Spatzen schießen" (Fritz Fleiner, Institutionen Des Deutschen Verwaltungsrecht, 8. Aufl., Tübingen, 1928). – user1551 Oct 06 '13 at 09:43
  • @anon could I ask about some of the details in this argument: what is the difference between $K(a_{ij},b_{ij})$ and $K[a_{ij},b_{ij}]$, I assumed that these were the same but then $K[a_{ij},b_{ij}] [T]$ should be all of $F[T]$ not a subring – T. Stark Sep 29 '19 at 13:43
  • @T.Stark Brackets means polynomial expressions in, parentheses means rational expressions in. – anon Dec 09 '19 at 20:46
  • If I understand correctly, we can always find a field such that $A,B$ are matrices in that field, and $A,B$ are invertible. Then we can cancel out $\det A$. However, I don't see why $A,B$ are invertilbe in $K(a_{ij}, b_{ij})$. Also, if this argument is correct, then we can also prove that $\det(AB) = \det(AC)$ implies $\det(B) = \det(C)$ by going to a field containing $A,B,C$ such that they are invertible. However, $\det(AB) = \det(AC) \implies \det(B) = \det(C)$ does not hold. There are plenty counter examples for $2\times 2$ matrices over $\mathbb{R}$ – TH Wang Sep 08 '20 at 08:00
  • @THWang $A$ and $B$ are invertible in $F$ because their inverses are explicitly given by Cramer's rule (the inverse is the adjugate matrix divided by the determinant - and the determinant is nonzero by noticing the terms of Leibniz expansion are distinct so no cancellation can occur). Note $a$s and $b$s are indeterminates, so this argument does not mean we can prove $A$ can be cancelled from $\det(AB)=\det(AC)$ when we fill the matrices with scalars. To illustrate, $ab=ac$ implies $u=v$ over the ring $K[a,b,c]$ where $a,b,c$ are indeterminates, but this doesn't hold when $a,b,c$ are scalars. – anon Apr 02 '22 at 02:02
  • I meant $ab=ac$ implies $b=c$ for indeterminates, not $u=v$. Typo introduced in revising comment. – anon Apr 02 '22 at 03:05
18

If $A$ is not invertible, $A_t=A+tI_n$ is invertible except for finitely many values of $t$. In particular, there is $\varepsilon > 0$ such that $A_t$ is invertible for every $t\in ]0,\varepsilon[$.

We therefore have $P(A_tB)=P(BA_t)$. Making $t\to 0$, we deduce $P(AB)=P(BA)$.

Community
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Ewan Delanoy
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  • If $A$ is not invertible, $A_t=A+tI_n$ is invertible except for finitely many values of $t$. In particular, there is $\varepsilon > 0$ such that $A_t$ is invertible for every $t\in ]0,\varepsilon[$. Is this because $Gl(n, R)$ is dense in $M(n,R)$? Thank you. – Matha Mota Nov 27 '20 at 08:05
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    @EmmyRahman This is because $\det(A_t)$ is a monic polynomial in $t$ with degree $n$, so it has at most $n$ roots. This also shows that $Gl(n,R)$ is dense in $M(n,R)$. – Ewan Delanoy Nov 27 '20 at 09:10
  • Thank you very much. I am sorry for this naive question, but I was stuck with this little thing. Thank you agian. – Matha Mota Nov 27 '20 at 09:26
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One way to do this is to note that $P(AB)-P(BA)$ is a polynomial in the entries of $A$ and $B$. You have shown that this vanishes for all invertible $A$, and want to show that it is precisely the zero polynomial. It suffices to note that the set of invertible $n\times n$ matrices over $\mathbb R$ or $\mathbb C$ is dense, and since polynomials are continuous it follows that $P(AB)-P(BA)$ is zero everywhere and thus the zero polynomial.

Alex Becker
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