MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A,B\in\mathbb{R}^{n\times n}$. Suppose that B is nonsingular and $AB\neq BA$. Can we always find real numbers $t_1,⋯,t_p$ such that

$$B\left(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\right)A=A\left(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\right)B?$$ N.B : $p$ is not fixed.

share|cite|improve this question
You should also link your other question… putting in evidence that if both $A$ and $B$ are non-singular, one can easily find counterexamples. – Valerio Capraro Jun 10 '12 at 10:12
up vote 12 down vote accepted

No. A counter-example is $$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 &0 &0 & 2 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & -2 & 0 \end{pmatrix}.$$

Observe that both matrices live in the image of the embeding $GL_2(\mathbb{C}) \to GL_4(\mathbb{R})$. In these terms, the matrices are $$A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad B = \begin{pmatrix} i & 1 \\ 0 & 2i \end{pmatrix}.$$

For any real $t$, the element $A+tB$ is of the form $\left( \begin{smallmatrix} u & v \\ 0 & 2u \end{smallmatrix} \right)$ for $u$ an nonzero complex number. So any product $\prod_{i=1}^p (A+t_i B)$ is of the form $\left( \begin{smallmatrix} u & v \\ 0 & 2^p u \end{smallmatrix} \right)$ for $u$ a nonzero complex number. Now $$A \begin{pmatrix} u & v \\ 0 & 2^p u \end{pmatrix} B = \begin{pmatrix} u & u + 2iv \\ 0 & 2^{p+1} i u \end{pmatrix}$$ $$B \begin{pmatrix} u & v \\ 0 & 2^p u \end{pmatrix} A = \begin{pmatrix} u & 2^{p+1} u + 2iv \\ 0 & 2^{p+1} i u \end{pmatrix}$$ So $2^{p+1} u = u$ and, as $u \neq 0$, we have a contradiction.

A vague attempt to explain how I found this. Let $C$ be the centralizer of $A^{-1}B $. The set of elements $M$ such that $AMB=BMA$ is $C A^{-1}$. The curve $A+tB$ is in $AC$ and it seems hard to find any properties of it which aren't true of all of $AC$. So we basically want to know whether $(AC)^p$ meets $CA^{-1}$ for sufficiently large $p$. I flirt with Hecke operators (because we are basically multiplying double cosets) and ergodic theory (because we want to know if a small subset of a group spreads out under multiplication).

Eventually I decide that the answer is probably "no". If so, then the most likely way to prove it is to find a function $f:GL_n(\mathbb{R}) \to \mathbb{R}$ which is more or less multipliciative, large for positive powers of $A$, small for negative powers of $A$ and bounded on $C$. A character would be great. But the only character of $GL_n$ is determinant and it doesn't work. Eventually, I realize that I can arrange for $A+tB$ to lie in a subgroup $H$ of $GL_n$ which has additional characters; let $T$ be the abelianization of $H$. It also seems like a good idea to make all of the eigenvalues of $A^{-1} B$ complex, as then $A+tB$ is always invertible. In other words, $T$ should have rank at least $2$, and should have lots of $\mathbb{C}^{\ast}$ factors. So I decide to try $H$ $$\begin{pmatrix} a & b & \ast & \ast \\ -b & a & \ast & \ast \\ 0 & 0 & c & -d \\ 0 & 0 & d & c \end{pmatrix}$$ with character $f=\frac{c^2+d^2}{a^2+b^2}$. By taking $f(A) = f(B)>1$, I ensure that $f(C)=1$, so $f$ is bounded on $C$ and large on $A$ as desired. At this point I stop trying to be systematic and just try some examples until one works.

share|cite|improve this answer

Your Answer


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.