Hello!

Let $m$ be an odd positive integer. Let $r$ be the smallest extension order of finite field $GF(q)$ that the $m$-th roots of unity are in $GF(q^r)$. For instance, it can be assumed that $q = 2$.

Let $\mathbf{d}_i$, $1 \le i \le m$ be unknown $k$-dim vectors over $GF(q)$. Let $G$ be given $k \times l$ matrix over $GF(q)$. Let $\mathbf{y}$ be unknown $l$-dim vector over $GF(q^r)$. Let $H$ be $s \times m$ matrix over $GF(q^r)$ of full rank over $GF(q^r)$.

I have to find unknown vectors $\mathbf{d}_i$, $1 \le i \le m$ and $\mathbf{y}$ so the following equation vanishes: $(\mathbf{d}_1^T G \mathbf{y}, \mathbf{d}_2^T G \mathbf{y}, \ldots, \mathbf{d}_m^T G \mathbf{y}) H^T = 0$. How it can be done?

Thank you very much!

Hello!

Let $m$ be an odd positive integer. Let $r$ be the smallest extension order of finite field $GF(q)$ that the $m$-th roots of unity are in $GF(q^r)$. For instance, it can be assumed that $q = 2$.

Let $\mathbf{d}_i$, $1 \le i \le m$ be unknown $k$-dim vectors over $GF(q)$. Let $G$ be given $k \times l$ matrix over $GF(q)$. Let $\mathbf{y}$ be unknown $l$-dim vector over $GF(q^r)$. Let $H$ be $s \times m$ matrix over $GF(q^r)$ of full rank over $GF(q^r)$.

I have to find unknown non-zero vectors $\mathbf{d}_i$, $1 \le i \le m$ and $\mathbf{y}$ so the following equation vanishes: $(\mathbf{d}_1^T G \mathbf{y}, \mathbf{d}_2^T G \mathbf{y}, \ldots, \mathbf{d}_m^T G \mathbf{y}) H^T = 0$. How it can be done?

Thank you very much!