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:


1. $A\Big(\displaystyle\prod\_{i=1}^{p}(A+t\_{i}B)\Big)B=B\Big(\displaystyle\prod\_{i=1}^{p}(A+t\_{i}B)\Big)A$
2. $\Big(\displaystyle\prod\_{i=1}^{p}(A+t\_{i}B)\Big)B\quad\mbox{is nonsingular and diagonalizable }$?


**N.B :**


1. The integer $p$ is not fixed.
2. 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 question about matrices with more details
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
$$A\prod (A+t\_iB)B=B,$$
$$B\prod (A+t\_iB)A=0.$$


