# 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:

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.

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Perhaps this question will get more attention when you provide a little bit more background. For example, what about the case $n=2$? What makes you think that there are such $t_i$? – Martin Brandenburg May 29 '12 at 9:23
Is $p$ fixed or variable? – Igor Rivin May 29 '12 at 14:15
It seems like there are two parts to this question: Understanding the set of $C$ such that $ACB=BCA$, and understanding the set of matrices which can be written as $\prod (A+t_i B)$. The first part isn't so hard. It is a linear space, of dimension at least $n$, and generically of dimension exactly $n$. For generic $(A,B)$, the space of possible $C$'s has basis $A^{-1} (B A^{-1})^k$, for $0 \leq k < n$. I haven't had much luck finding a way to think about the second question. – David Speyer May 29 '12 at 15:10
I just ran the following quick experiment: I generated two random $2 \times 3$ matrices (namely, {{61, 82, 81}, {99, 0, 82}, {24, 67, 11}} and {{28, 55, 16}, {63, 59, 68}, {84, 76, 35}}) and solved the linear equation $A (p A^2 + q AB + r BA + s B^2) B = B (p A^2 + q AB + r BA + s B^2) A$ for $(p,q,r,s)$. There were no nonzero roots. So, if there is a formula like the above, the formula for $C$ must have degree $>2$. – David Speyer May 29 '12 at 18:22
It is not a good idea to edit your original question so that the answers posted before don't make any sense. You should have accepted the correct answer and start a new question, I think. – Vladimir Dotsenko Jun 1 '12 at 8:51

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 $$A\prod (A+t_iB)B=B,$$ $$B\prod (A+t_iB)A=0.$$

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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 $0$.

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