A question about matrices with more details - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:46:37Z http://mathoverflow.net/feeds/question/98260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98260/a-question-about-matrices-with-more-details A question about matrices with more details driss-alamilouati 2012-05-29T08:28:28Z 2012-05-31T19:28:17Z <p>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 $$B^{-1}A=\begin{pmatrix} \lambda_{1}&amp; 1&amp;&amp;&amp;&amp;&amp;&amp;\cr &amp;\lambda_{1}&amp;\ddots&amp;&amp;&amp;&amp;&amp;\cr &amp;&amp;\ddots&amp;&amp;&amp;\LARGE{0}&amp;&amp;\cr &amp;&amp;&amp;\lambda_{r}&amp;1&amp;&amp;&amp;\cr &amp;&amp;&amp;&amp;\lambda_{r}&amp;&amp;&amp;\cr &amp;&amp;&amp;&amp;&amp;\lambda_{r+1}&amp;&amp;\cr &amp;&amp;\LARGE{0}&amp;&amp;&amp;&amp;\ddots&amp;\cr &amp;&amp;&amp;&amp;&amp;&amp;&amp;\lambda_{m} \end{pmatrix}$$ (<em>It is the canonical Jordan form</em>) Can we ever find reals numbers $t_ {1}, \cdots, t_ {p}$ so that the two following assertions are true:</p> <ol> <li>$A\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B=B\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)A$</li> <li>$\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B\quad\mbox{is nonsingular and diagonalizable }$?</li> </ol> <p><strong>N.B :</strong></p> <ol> <li>The integer $p$ is not fixed.</li> <li>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.</li> </ol> <p>Thanks for help.</p> http://mathoverflow.net/questions/98260/a-question-about-matrices-with-more-details/98304#98304 Answer by Michael Renardy for A question about matrices with more details Michael Renardy 2012-05-29T20:30:28Z 2012-05-29T20:30:28Z <p>Let $$A=\pmatrix{1&amp;0\cr 0&amp;0},\quad B=\pmatrix{0&amp;1\cr 0&amp;0}.$$ Then $A^2=A$, $AB=B$, $BA=0$, $B^2=0$. It follows that $$A\prod (A+t_iB)B=B,$$ $$B\prod (A+t_iB)A=0.$$</p> http://mathoverflow.net/questions/98260/a-question-about-matrices-with-more-details/98351#98351 Answer by Denis Serre for A question about matrices with more details Denis Serre 2012-05-30T10:02:28Z 2012-05-30T11:19:39Z <p>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 $$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\prod_{j=1}^p(s_jA+t_jB)\right)A.$$</p> <p>If you pose the question in an algebraically closed field $k$ (say, $k=\mathbb C$), then the answer is <strong>yes</strong> for the following reason:</p> <blockquote> <p>There exist $2^n-1$ non-zero factors $s_jA+t_jB$ such that $\prod_{j=1}^n(s_jA+t_jB)=0$.</p> </blockquote> <p>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 $0$.</p>