A question about matrices with more details Let $A,B\in\mathbb{R}^{n\times n}$. Suppose that $B$ is nonsingular and that  there exists  $m$ reals pairwise distinct  $\lambda_{1},\cdots,\lambda_{r},\cdots,\lambda_{m}$  such that 
$$B^{-1}A=\begin{pmatrix}
\lambda_{1}& 1&&&&&&\cr
&\lambda_{1}&\ddots&&&&&\cr
&&\ddots&&&\LARGE{0}&&\cr
&&&\lambda_{r}&1&&&\cr
&&&&\lambda_{r}&&&\cr
&&&&&\lambda_{r+1}&&\cr
&&\LARGE{0}&&&&\ddots&\cr
&&&&&&&\lambda_{m}
\end{pmatrix}$$
(It is the canonical Jordan form)
Can we ever find reals numbers $ t_ {1}, \cdots, t_ {p} $ so that the two following assertions are true:


*

*$A\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B=B\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)A$

*$\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B\quad\mbox{is nonsingular and diagonalizable }$?


N.B :


*

*The integer $p$ is not fixed.

*This question has arisen when studying the contollability of a real discrete-time nonlinear system. This explains why the matrices are supposed to be reals.


Thanks for  help.
 A: To my taste, it seems more natural to let $A$ and $B$ play symmetric role, by asking whether there exists non-trivial factors $s_jA+t_jB$ such that 
$$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\prod_{j=1}^p(s_jA+t_jB)\right)A.$$
If you pose the question in an algebraically closed field $k$ (say, $k=\mathbb C$), then the answer is yes for the following reason:

There exist $2^n-1$ non-zero factors $s_jA+t_jB$ such that $\prod_{j=1}^n(s_jA+t_jB)=0$.

The proof is by induction over the rank of products $\prod_{j=1}^p(s_jA+t_jB)=0$. Suppose that exists such a product $\Pi$, with rank $r\ge1$. Let us write
$$\Pi=\sum_{j=1}^rx_ja_j^T.$$
Then
$$\Pi M\Pi=\sum_{i,j=1}^r(a_i^TMx_j)x_ia_j^T.$$
The rank of $\Pi M\Pi$ will be less than or equal to $r-1$ if $\det(a_i^TMx_j)_{1\le i,j\le r}=0$. When $M=sA+tB$, this writes $H(s,t)=0$ where $H$ is a homogeneous polynomial of degree $r$. If $r\ge1$, it does have a non-trivial zero. Then $\Pi':=\Pi(sA+tB)\Pi$ is an other product, with rank $\le r-1$. If in addition $\Pi$ has $2^{n-r}-1$ factors, then $\Pi'$ has $2^{n+1-r}-1$ factors. After $n$ steps, one obtains a product of $2^n-1$ factors whose rank is $0$.
A: Let
$$A=\pmatrix{1&0\cr 0&0},\quad B=\pmatrix{0&1\cr 0&0}.$$
Then $A^2=A$, $AB=B$, $BA=0$, $B^2=0$. It follows that
$$A\prod (A+t_iB)B=B,$$
$$B\prod (A+t_iB)A=0.$$
