Is a simple graph matrix the sum of a "shiftordered" matrix and its transposed matrix

Question

This is the generalization of a question Is a simple graph the "sum" of a partial order and its dual?

Nik Weaver found a counterexample in a very nice, complete (and instantaneous!) answer, so I modify the question to hope for a yes.

(I changed the old $S$ of previous question in $G$ in this question )

A "$n$-graph matrix" $G\in M_n(\mathbb F_2)$ is a symmetric matrix such that $G_{ii}=0$ for all $i\in [1,n]$

A "$n$-order matrix" $T\in M_n(\mathbb F_2)$ is a matrix such that there exists a partial ordered relation $\leq_T\subset [1,n]^2$ such that :

$T_{ij}=1\Leftrightarrow i\leq_T j$

Let's add $m\in \mathbb N$ full zero first rows and $m$ full zero last columns to an $n$-order matrix then we will call it a $(m+n)$-schiftordered matrix. (since $m$ can be $0$, an ordered matrix is a schiftordered matrix)

Is it true that for any "$n$-graph matrix" $G$ there exists a "$n$-schiftordered matrix" $Z$ s.t. $G=Z+Z^t$ ?

(Note that if $S$ is the one that occures in Nik Weaver answer then $Z$ defined by $Z_{ij}=S_{ij}$ if $i>j$ and $0$ if not, satisfies $Z+Z^t=S$ and $Z$ is a $5$ -shiftordered matrix)

Answer 1

Expanded comment: It seems that essentially the same counting argument applies as the one for the previous question (mathoverflow.net/q/306562):

We can make first $m$ rows and last $m$ columns of an $n\times n$ matrix zero in $n+1$ different ways, this increases the total number of matrices $S$ at most $(n+1)$ times.