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.

  • 4
    $\begingroup$ 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 $t_i$? $\endgroup$ – Martin Brandenburg May 29 '12 at 9:23
  • 3
    $\begingroup$ Is $p$ fixed or variable? $\endgroup$ – Igor Rivin May 29 '12 at 14:15
  • 6
    $\begingroup$ 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. $\endgroup$ – David E Speyer May 29 '12 at 15:10
  • $\begingroup$ 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 $>2$. $\endgroup$ – David E Speyer May 29 '12 at 18:22
  • 2
    $\begingroup$ 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. $\endgroup$ – 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 $$A\prod (A+t_iB)B=B,$$ $$B\prod (A+t_iB)A=0.$$

| cite | improve this answer | |

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 $0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.