existence of polynomial equation system solution - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:31:57Z http://mathoverflow.net/feeds/question/86695 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86695/existence-of-polynomial-equation-system-solution existence of polynomial equation system solution Seyong 2012-01-26T07:16:23Z 2012-03-23T09:22:01Z <p>For $1 \leq i \leq n$, let <code>$A=\begin{bmatrix} a_{11} &amp; \cdots &amp; a_{1n} \\ \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; \cdots &amp; a_{nn} \\ \end{bmatrix}$</code></p> <p><code>$B_i=\begin{bmatrix} b_{i1} \\ \vdots \\ b_{in} \end{bmatrix}$</code> and <code>$C_i=\begin{bmatrix} c_{i1} \\ \vdots \\ c_{in} \end{bmatrix}^*$</code></p> <p>Let $D=A+\sum_{1 \leq i \leq n}B_i k_i C_i$. Then, for almost all $a_{ij},b_{ij},c_{ij}$, there exists $k_i \in \mathbb{C}$ such that all eigenvalues of $D$ are zeros.</p> http://mathoverflow.net/questions/86695/existence-of-polynomial-equation-system-solution/86756#86756 Answer by Gottfried Helms for existence of polynomial equation system solution Gottfried Helms 2012-01-26T21:20:16Z 2012-01-27T07:33:15Z <p><em>[update] By an example of 4x4-matrices the ansatz below could not be used to solve the problem. The matrix</em> $\small Q_K$ <em>cannot in general be made lower triangular by choices of the</em> $\small k_i$. <em>I'll delete this answer soon if I cannot improve the ansatz</em>.<br> <hr> I do not yet have a full answer but possibly a first step into one. I think that this can be answered unsing two facts: </p> <p>a) There is a similarity transformation with a rotation <strong><em>T</em></strong> such that $\small P = T^{-1}\cdot A \cdot T = T' \cdot A \cdot T$ where <strong><em>P</em></strong> is triangular and has the eigenvalues of <strong><em>A</em></strong> on its diagonal. </p> <p>b) Your sum-expression of vector outer products, let it be called matrix $\small E_K = \sum_{i=1}^n B_i k_i C_i = \sum_{i=1}^n k_i (B_i C_i)= \sum_{i=1}^n k_i E_i$ is a weighted (by the $\small k_i$ weights) sum of rank-<em>1</em>-matrices $\small E_i$ </p> <p>From the "similarity rotated version" of all matrices </p> <p>$\qquad \small Q_K = T' E_K T = \sum_{i=1}^n k_i (T' E_i T) = \sum_{i=1}^n k_i Q_i$ (which should be made triangular by choices of $\small k_i$ ) and<br> $\qquad \small R = T' D T$ which is then also triangular </p> <p>we get your final equation in its form with triangular matrices </p> <p>$\qquad \small R = P + Q_K$ </p> <p>We'll have a solution if the weights $\small k_i$ for the <strong><em>non-triangular</em></strong>, generic but <strong><em>rank-1</em></strong>-matrices $\small Q_i$ can be chosen such that their sum $\small Q_K$ becomes triangular <strong><em>and</em></strong> its diagonal equals the negative diagonal in $\small P$. </p> <p>I've a vague speculation that the equation with this triangular matrices can easier be shown to be "almost always" possible, but have not yet a further concrete approach. At least this construction exhibits that the rank-1-matrices $\small E_i$ (and thus $\small Q_i$ ) cannot be simple scalar multiples of each other if <strong><em>A</em></strong> has full rank and thus the restriction to "for almost all" is unavoidable and possibly mainly consists of this property.</p>