Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given two matrices $A,B.$ On what conditions does $AB \sim BA$ hold?

share|improve this question
1  
Since it certainly holds if $A$ and $B$ commute, it would be nice to see an example where $AB\ne BA$ yet $AB \sim BA$. –  Greg Martin Jul 28 '12 at 0:01
    
Is is possible to prove that both share the same eigenvalues? –  Tomer Galanti Jul 28 '12 at 0:03

4 Answers 4

up vote 5 down vote accepted

If $A$ is invertible, then $AB = A(BA)A^{-1}$ which shows that $AB$ and $BA$ are similar. Similar (no pun intended) proof if $B$ is invertible.

share|improve this answer

$AB$ is conjugate to $BA$ if either $A$ or $B$ are invertible. If neither is the case, there are counterexamples: for example, it may be the case that $AB = 0$ while $BA \neq 0$. Explicitly, take $$A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right].$$

We have $AB = A$ but $BA = 0$.

However, there is a salvage: $AB$ and $BA$ have the same characteristic polynomial. See this blog post. (Short proof: this must hold if either $A$ or $B$ is invertible, and that condition is Zariski dense.)

share|improve this answer
    
Can you please provide with a reference regarding the case one of the matrices are invertiable ? –  Belgi Jul 28 '12 at 0:51
2  
@Belgi If $A$ is invertible, and if $C=A^{-1}$, then $C(AB)C^{-1}=BA$ as you can check. Similarly, if $B$ is invertible, and if $C=B^{-1}$, then $C(BA)C^{-1}=AB$. –  Amitesh Datta Jul 28 '12 at 1:28

As mentioned already, if either of $A$ or $B$ is invertible (and both are the same size), we have $$ \begin{align} AB=A(BA)A^{-1}\quad&\mbox{if $A$ is invertible}\\ AB=B^{-1}(BA)B\quad&\mbox{if $B$ is invertible} \end{align} $$

However, here is a short proof that even if $A$ is $m\times n$ and $B$ is $n\times m$, the characteristic polynomials of $AB$ and $BA$ differ only by a factor of $\lambda^{\large|n-m|}$.

share|improve this answer

The proof is rather straightforward and I do not assume anything about $A$ and $B$, except that they are square.

I do so by showing that they have the same eigenvalues. If that is the case then $A=MJM^{-1}$ and $B=CJC^{-1}$ both share the same $J$ in their Jordan form and I take for granted that there exist some matrix $K$ such that $C=KM$, in order to have $B=KMJ(KM)^{-1}$.

So suppose $\lambda$ is an eigenvalue of $AB$. Then $ABx=\lambda x$.

Premultiplying both sides by $B$, we get $BABx =BA(Bx) \lambda(Bx)$, which shows that $\lambda$ is also an eigenvalue of $BA$.

share|improve this answer
1  
Having the same eigenvalues doesn't mean they share a Jordan form. You should read the other answers, the result is false if $A$ and $B$ are non-invertible. –  EuYu Nov 26 '12 at 0:42
    
You're right, Ive just seen the counter example. I wrote it thinking they had a full set of eigenvalues, in which case that should be true of J. –  Michael P. Nov 26 '12 at 0:55
    
Please note in my edit: the singular form of "matrices" is "matrix." –  rschwieb Nov 26 '12 at 1:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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