We prove that the maximal dimension of a subspace $V$ of the generic tensor product of $m$ symbol algebras of prime degree $p$ with $\operatorname{Tr}(v^{p-1})=0$ for all $v\in V$ is $\frac{p^{2m}-1}{p-1}$. The same upper bound is thus obtained for $V$ with $\operatorname{Tr}(v)=\operatorname{Tr}(v^2)=\dots=\operatorname{Tr}(v^{p-1})=0$ for all $v \in V$. We make use of the fact that for any subset $S$ of $\underbrace{\mathbb{F}_p \times \dots \times \mathbb{F}_p}_{n \ \text{times}}$ of $|S| > \frac{p^{n}-1}{p-1}$, for all $u\in V$ there exist $v,w\in S$ and $k\in [\![0,p-1]\!]$ such that $kv+(p-1-k)w=u$.
