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](https://arxiv.org/abs/2003.10529) 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 $\emptyset \neq T \subseteq S$, 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](https://arxiv.org/abs/1812.00264)).



> 
> **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 $M\_1 \circ \dots \circ M\_m$ is disconnected.
> 
> 
> 


I have proven this conjecture when $m=2$; or $m=3$ and $r\_3=2$; or $m$ is arbitrary, $r\_1\geq 1$ is arbitrary, and $r\_2=\dots=r\_m=2$.



A Hadamard product of binary (or ternary) matroids
I hope that below is the proof of Conjecture (for $n\geqslant 2$, for $n=1$ it is false by trivial reasons), but please check carefully.


If $M$, $N$ are matroids on the ground set $E$ which are represented over a field $\mathbb{F}$: $M=\{x\_i:i\in E\}$, $N=\{y\_i:i\in E\}$ ($x\_i$ and $y\_i$ are vectors in corresponding vector spaces $X$, $Y$ over $\mathbb{F}$), we define $M\circ N$ as a matroid on the ground set $E$ over $\mathbb{F}$ corresponding to the tensor products $\{x\_i\otimes y\_i:i\in E\}$.


Conjecture follows immediately from the following two lemmata.


**Lemma 1.** If $N$ or $M$ is disconnected, then so is $M\circ N$.


**Proof.** Obvious.


**Lemma 2.** If $N$, $M$ and $M\circ N$ are connected, then ${\rm rank} (M\circ N)\geqslant {\rm rank}(M)+{\rm rank}(N)-1$.


**Proof.** We use the ear decomposition of the connected matroid $M\circ N$: $E$ may be represented as a union of circuits $E=C\_1\cup C\_2\cup \ldots \cup C\_m$ so that the sets $E\_k:=C\_1\cup \ldots \cup C\_k$ satisfy $C\_k\cap E\_{k-1}\ne \emptyset$, ${\rm rank}(E\_k)-{\rm rank}(E\_{k-1})=|E\_k\setminus E\_{k-1}|-1$ for all $k=2,\ldots,m$ (hereafter ${\rm rank}$, ${\rm rank}\_M$, ${\rm rank}\_N$ denote rank functions of $M\circ N$, $M$, $N$ respectively).


For completeness, I prove that it exists. We start with arbitrary circuit $C\_1$. Assume that $C\_1,\ldots,C\_k$ are already constructed. Let $B$ be a base of $E\_k$. Choose consequently independent elements $f\_1,f\_2,\ldots$ in $E\setminus E\_k$ until we get ${\rm rank}(B\cup \{f\_1,f\_2,\ldots,f\_r\})<r+|B|$. If this never happens, we get ${\rm rank}(E\_k)+{\rm rank}(E\setminus E\_k)={\rm rank}(E)$, and $E$ is disconnected. So sometimes it happens. The set $B\cup\{f\_1,\ldots,f\_{r}\}$ is dependent, but if we remove $f\_r$ it becomes independent. So it contains a unique circuit (containing $f\_r$), which may serve as $C\_{k+1}$. Proceed this way.


Now the inequality ${\rm rank}(E)\geqslant {\rm rank}(M)+{\rm rank}(N)-1$ follows from the following two propositions:


1. $|E\_1|-1={\rm rank}(C\_1)\geqslant {\rm rank}\_{M}(C\_1)+{\rm rank}\_N(C\_1)-1$;
2. For each $k\geqslant 2$, $$|E\_k\setminus E\_{k-1}|-1\geqslant {\rm rank}\_{M}(E\_k)-{\rm rank}\_{M}(E\_{k-1})+{\rm rank}\_{N}(E\_k)-{\rm rank}\_{N}(E\_{k-1}).$$


We proceed with proving 1). Denote $C\_1=\{1,2,\ldots,r\}$; ${\rm rank}\_{M}(C\_1)=\alpha$; let $x\_1,\ldots,x\_{\alpha}$ be an $M$-base of $C\_1$. Assume that there exists $i\in \{1,2,\ldots,\alpha\}$ for which $y\_i$ does not belong to a span of $y\_{\alpha+1},\ldots,y\_r$. Then there exists linear functionals $\eta,\nu$ on $X$, $Y$ respectively satisfying $\eta(x\_i)=\nu(y\_i)=1$, $\eta(x\_j)=0$ for $j\in \{1,\ldots,\alpha\}\setminus \{i\}$, $\nu(y\_j)=0$ for $j=\alpha+1,\ldots,r$. Then
$\eta\otimes \mu$ applied to $x\_j\otimes y\_j$ equals 1 for $j=1$ and equals 0 for $j\in \{1,\ldots,\alpha\}\setminus \{i\}$. Thus $x\_i\otimes y\_i$ can not be a linear combination of other elements of our circuit, a contradiction. Therefore $y\_{\alpha+1},\ldots,y\_r$ span $y\_1,\ldots,y\_r$ and ${\rm rank}\_N (C\_1)\leqslant r-\alpha=|C\_1|-{\rm rank}\_{M}(C\_1)$ that is 1).


The final step is 2). Denote $C\_k\setminus E\_{k-1}=\{1,\ldots,r\}. $Let $x\_1,\ldots,x\_{\alpha}$ form a basis of $C\_{k}\setminus E\_{k-1}$ in the matroid $M/E\_{k-1}$ (that corresponds to factorizing modulo the span of $\{x\_i:i\in E\_{k-1}\}$.) So, $\alpha={\rm rank}\_{M}(E\_k)-{\rm rank}\_{M}(E\_{k-1})$. Again, assume that there exists $i\in \{1,2,\ldots,\alpha\}$ for which $y\_i$ does not belong to a span of $y\_{\alpha+1},\ldots,y\_r$. Then there exists linear functionals $\eta,\nu$ on $X$, $Y$ respectively satisfying $\eta(x\_i)=\nu(y\_i)=1$, $\eta(x\_j)=0$ for $j\in E\_{k-1}\cup\{1,\ldots,\alpha\}\setminus \{i\}$ , $\nu(y\_j)=0$ for $j=\alpha+1,\ldots,r$. We get a contradiction similarly to the proof of 1). Therefore ${\rm rank}\_N(C\_k\setminus E\_{k-1})\leqslant r-\alpha$. But by Lemma 1 the matroid induced by $N$ on $C\_{k}$ is connected, thus (the first inequality is a submodularity of the rank function) we get $${\rm rank}\_N(E\_k)-{\rm rank}\_N(E\_{k-1})\leqslant {\rm rank}\_N(C\_k)-{\rm rank}\_N(C\_k\cap E\_{k-1})\\ \leqslant {\rm rank}\_N(C\_k\setminus E\_{k-1})-1\\
\leqslant r-\alpha-1=|E\_k\setminus E\_{k-1}|-({\rm rank}\_{M}(E\_k)-{\rm rank}\_{M}(E\_{k-1}))-1$$
that proves 2).


