We determine that the maximum rank of an order-n $(\ge 2)$ tensor with format $2\times \cdots \times 2$ over the finite field ${\mathbb{F}}_2$ is $2\cdot 3^{n/2-1}$ for even n, and $3^{\lfloor n/2\rfloor }$ for odd n. Since tensor rank is non-increasing upon taking field extensions, ${\mathbb{F}}_2$ gives the largest rank attainable for this tensor format. We also determine a maximum rank canonical form and compute its orbit under the action of the symmetry group $G{L}_2({\mathbb{F}}_2{)}^{\times n}$, and prove that this is the unique maximum rank canonical form, for even $n\ge 2$.

中文翻译：

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rank2上2×⋯×2张量的最大秩

我们确定订单的最大等级-n $（\ge 2）$ 张量与格式 $2\times \cdots \times 2$ 在有限域上 ${\mathbb{F}}_2$ 是 $2\cdot 3^{ñ/2-\mathrm{1个}}$甚至ñ，和$3^{\lfloor ñ/2\rfloor }$为奇数n。由于张量等级在进行场扩展时不会增加，${\mathbb{F}}_2$给出此张量格式可获得的最大等级。我们还确定最大秩规范形式，并在对称群​​的作用下计算其轨道$G{\mathrm{大号}}_2（{\mathbb{F}}_2{）}^{\times ñ}$，并证明这是唯一的最大秩规范形式，即使 $ñ\ge 2$。