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 $B$.
$$C=\pmatrix{c\_1& c\_2\\c\_3&c\_4}=\pmatrix{a\_1b\_1& a\_2b\_2\\a\_3b\_3&a\_4b\_4}=\pmatrix{a\_1& a\_2\\a\_3&a\_4} \circ \pmatrix{b\_1& b\_2\\b\_3&b\_4}=A \circ B.$$
$$c\_1=a\_1b\_1,\quad c\_2=a\_2b\_2,\quad c\_3=a\_3b\_3,\quad c\_4=a\_4b\_4.$$
$$T:\mathcal{A} \times\ \mathcal{B} \to \mathcal{C}$$
where $T\in \mathcal{A}^\* \otimes \mathcal{B}^\* \otimes \mathcal{C}$,
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 express this product as a sum of elementary tensors:
$$T=\sum^{4}\_{i=1} \ a\_i\otimes b\_i\otimes c\_i$$
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. 



Strassen-like algorithm for Hadamard product of $2\times 2$ matrices
Let's write $T \in \mathcal{A}^\* \otimes \mathcal{B}^\* \otimes \mathcal{C}$ for the tensor, and $L\_T : \mathcal{A} \otimes \mathcal{B} \to \mathcal{C}$ for the linear map given by $A \otimes B \mapsto A \circ B$. This $L\_T$ is one of the flattenings of $T$.


You gave an expression for $T$ as a sum of four simple tensors, so the tensor rank satisfies $\operatorname{rank}(T) \leq 4$.


On the other hand, $\operatorname{rank}(T) \geq \operatorname{rank}(L\_T)$. (This is the "flattening bound for rank".) One way to see this inequality is that if $U$ is a simple tensor, then $L\_U$ has rank $1$; and $L\_{U+V} = L\_U + L\_V$; so if $T$ is a sum of $r$ simple tensors, then $L\_T$ is a sum of $r$ rank-one maps, so $\operatorname{rank}(L\_T) \leq r$.


Finally, for $T$ and $L\_T$ as above, we can find the rank of $L\_T$. It has rank $4$ because $L\_T$ is surjective. Given any $2 \times 2$ matrix $C$, we can get $C$ as the image of, say, $A \circ B = C$ where $A=C$ and $B$ is the matrix of all $1$s.


So $\operatorname{rank}(T) \geq \operatorname{rank}(L\_T) = 4$.


