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Peter Mueller provided a negative answer for $n=4$. Based on it, we show that the answer is negative for any even $n\ge 4$. Indeed, suppose for a contradiction that the space $V=\mathbb R^n$ admits a …

This is a draft proof of an affirmative answer to Problem 3. Proposition. For any $n\in\mathbb N$, $\varepsilon>0$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $ …

To mark the question answered, I copied below my accepted answer from Mathematics Stack Exchange. There are no matrices $M_1$ and $M_2$ which you are trying to find because of the following propositio …

The following argument suggests that the factor $O(n)$ cannot be improved. Suppose for a contradiction that $x\in\{0,1\}^n$ and after queries with $v_1,\dots,v_{n-1}\in\mathbb R^n$ we determined that …

There is a special version of this question (for $n=15$ and $m\le 7$) at Mathematics.SE. For each $n\ge 2$ I constructed a set $S$ with the property requiring $m\ge \left\lfloor\tfrac n2\right\rfloor= …

Too long for a comment. We can simplify the proposed inequality as follows. Put $F=\sum_{i=1}^n f_i^2$, $G=\sum_{i=1}^n g_i^2$, and $H=\sum_{i=1}^n f_ig_i$. Remark that $H\ge 0$ and $$\|\nabla f …

Because of this symmetry, it is said that one can just fix one of the values of the indices, say $i=0,1,..,b-1$ and generate the other elements from the symmetry relation above. The issue is very …