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.



construction of matrices verifying an identity
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.




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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.


