# Rudi Pendavingh

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## Registered User

 Name Rudi Pendavingh Member for 6 months Seen May 8 at 9:51 Website Location Netherlands Age 45
 Dec12 comment How to solve a system of quadratic equations over finite fields?If you write your vectors $d_i$ as the rows of a matrix $D$, then unless I misread something your problem is to find, given matrices $G$ and $H$, a vector $y$ and a matrix $D$ such that $$HDGy=0.$$ I guess you want $D,y$ nonzero? Or do you want all such $D,y$. Dec9 revised Convex optimization problem to QPPadded 4 characters in body Dec8 revised Convex optimization problem to QPPdeleted 2 characters in body Dec8 revised Convex optimization problem to QPPedited body Dec8 revised Convex optimization problem to QPPadded 754 characters in body Dec8 awarded ● Commentator Dec8 comment Convex optimization problem to QPPAdding $F_i(x)\geq 0$ will introduce restrictions on $x$ that are not in your original problem. Dec8 revised Convex optimization problem to QPPdeleted 16 characters in body Dec8 answered Convex optimization problem to QPP Dec8 awarded ● Enthusiast Dec5 comment linear independence of finite binary sequencesDo you really want your collection of vectors $\{x_1,\ldots, x_n\}$' to be unordered? Nov26 accepted A certain type of quadratic problem. Nov23 comment A certain type of quadratic problem.Actually, you will have $\{u\in Uâˆ£ u^HAu=0}\neq \emptyset$ for all linear combinations $A$ of $A_,\ldots,A_k$. But still that is not enough to force a common point I guess. Nov23 comment A certain type of quadratic problem.I expect that there are examples with two constraints $u^HA_iu=0$ where there is a gap. My argument to close the gap really breaks when there are more constraints. Each set $\{u\in U\mid u^HA_iu=0\}$` will be nonempty, but there need not be a common point to all these sets. Nov22 comment A certain type of quadratic problem.Well, if you have a method at your disposal for computing the smallest eigenvalue of a Hermitian matrix, then you can evaluate $\lambda(t)$ for any $t$. As $t\mapsto\lambda(t)$ is concave, solving $\max_t \lambda(t)$ can be done by standard methods, say bisection. Nov22 revised A certain type of quadratic problem.deleted 1 characters in body Nov22 answered A certain type of quadratic problem.