Let $A=\pmatrix{a_1& a_2\\a_3&a_4}, B=\pmatrix{b_1& b_2\\b_3&b_4}$ be two matrices. Let $C$ be the Hadamard product of $A$ and $B$. $$C=\pmatrix{c_1& c_2\\c_3&c_4}=\pmatrix{a_1b_1& a_2b_2\\a_3b_3&a_4b_4}=\pmatrix{a_1& a_2\\a_3&a_4} \circ \pmatrix{b_1& b_2\\b_3&b_4}=A \circ B.$$ $$c_1=a_1b_1,\quad c_2=a_2b_2,\quad c_3=a_3b_3,\quad c_4=a_4b_4.$$ $$T:A^* \times\ B^*\to C$$$$T:\mathcal{A} \times\ \mathcal{B} \to \mathcal{C}$$ where $T\in A^*\otimes B^*\otimes C .$$T\in \mathcal{A}^* \otimes \mathcal{B}^* \otimes \mathcal{C}$, where $\mathcal{A}=\mathcal{B}=\mathcal{C}=\operatorname{Mat}_{2 \times 2}(\Bbbk)$, the space of $2 \times 2$ matrices over the field $\Bbbk$. We can showexpress this product byas a sum of elementary Tensorstensors: $$T=\sum^{4}_{i=1}\alpha_{i,j,k}\ a_i\otimes b_i\otimes c_i$$$$T=\sum^{4}_{i=1} \ a_i\otimes b_i\otimes c_i$$ where the $a_i$, etc., correspond to standard bases for the spaces of matrices (i.e., the elementary matrices) (or the dual bases). For sure the rank of $T$ is at most $4$. I am trying to show that the rank is exactly $4$. I tried to find the rank by flattening but could not find any good flattening to give me the exact rank. Any idea about flattening? thanks.
Federico Poloni
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