I would like to know if anyone has studied the following “Hadamard product” of binary (or ternary) matroids. (There is a notion of Hadamard product of matroids studied e.g. here but I think that one is different.)

Let $M,N$ be simple binary matroids of rank $r$ and $s$, respectively, over the same ground set $E$ of size $n$. For binary representations $(x_1,dots, x_n)$ and $(y_1,dots, y_n)$ of $M$ and $N$, respectively, define the *Hadamard product* of $M circ N$ to be the binary matroid represented by $(x_1 otimes y_1, dots, x_n otimes y_n)$. One can easily show that this is a well-defined matroid product, using the fact all representations of binary matroids are projectively equivalent (Proposition 6.6.5, *Matroid Theory*, Oxley).

After a little work, one can derive the linearly independent sets in $M circ N$. Suppose WLOG that $(x_1,dots, x_r)$ form a basis for $M$. For each $i in {1,dots, r}$, let

$$text{Supp}(i)={a in {1,dots, n} | x_a(i) neq 0},$$

where $x_a(i)in mathbb{F}_2$ is the $i$-th coordinate of $x_a$ in the basis $(x_1,dots, x_r)$. Then $S subseteq (n)$ is linearly independent in $M circ N$ if and only if for all $T subseteq (n)$ of size $1 leq |{T}| leq n-1$, there exists $i in {1,dots, r}$ such that $sum_{a in T} x_a(i) y_a neq 0$. This inequality is equivalent (over $mathbb{F}_2$), to saying that the set

$$T cap text{Supp}(i)$$ is not Eulerian in $N$, i.e. it cannot be partitioned into circuits in $N$.

As a side note, I would also be very interested in any feedback on the following conjecture, which is the $mathbb{F}_2$-version of a conjecture I have been thinking about for some time (preprint here).

**Conjecture:** Let $M_1,dots, M_m$ be simple binary matroids of rank $r_1,dots, r_m$, respectively over the same ground set $E$ of size $n$. If $n leq sum_{j=1}^m (r_j-1)+1$, then $M_1 circ dots circ M_m$ does not form a circuit.

I have proven this conjecture when $m=2$; or $m=3$ and $r_3=2$; or $m$ is arbitrary, $r_1geq 1$ is arbitrary, and $r_2=dots=r_m=2$.