Dtjytyk

Consider the $3times 3$ matrix
$$
Aequiv begin{pmatrix}
mu_1-mu_1'& mu_1-mu_1'-c & mu_1-mu_1'-c-d\
mu_1+a-mu_1'& mu_1+a-mu_1'-c & mu_1+a-mu_1'-c-d\
mu_1+a+b-mu_1'& mu_1+a+b-mu_1'-c & mu_1+a+b-mu_1'-c-d\
end{pmatrix}
$$
where $mu_1, mu_1', a, b, c, d$ are real numbers and $a,b,c,d$ are strictly positive.

Notice that
$$
R_2=R_1+a\
R_3=R_1+a+b\
C_2=C_1-c\
C_3=C_1-c-d
$$
where $R_1,R_2,R_3,C_1,C_2,C_3$ are respectively the three columns and the 3 rows of $A$.

Claim that I have already shown: Assume

(1) $aneq b,cneq d$

(2) $mu_1,mu_1',a,b,c,d$ are such that, when ordering the elements of $A$ from smallest to largest, the obtained ordered sequence of points is symmetric around zero.
E.g., suppose that $mu_1,mu_1',a,b,c,d$ are such that $A$ contains only $3$ distinct elements which we denote by $eta_1<eta_2<eta_3$. Then, assumption (2) states that
$
eta_1=-eta_3$ and $
eta_2=0$.

Then $mu_1=mu_1', a=c, b=d$. In other words, $C_1=-R_1$,$C_2=-R_2$, $C_3=-R_3$.

Question: can we rewrite $(1)$ [or state sufficient conditions for $(1)$] in a way that involves some "well known" operations on the matrix $A$? E.g., as a sort of rank condition? Or as a sort of determinant condition? Or as a sort of linear independence condition?

More details (not sure they are needed)

I have developed the proof of my claim above in a "very constructive" way which somehow prevents me to understand if there is any " mathematically deeper" meaning of assumption (1) that can be stated in terms of the properties of the matrix $A$.

Let me report here a summary of the proof of my claim above.

1) First of all notice that we can establish a partial order of the elements of $A$, i.e.
$$
Aequiv begin{pmatrix}
mu_1-mu_1'&<& mu_1-mu_1'-c &<& mu_1-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a-mu_1'&<& mu_1+a-mu_1'-c &<& mu_1+a-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a+b-mu_1'&<& mu_1+a+b-mu_1'-c &<& mu_1+a+b-mu_1'-c-d\
end{pmatrix}
$$

2) Let $eta_1<eta_2<...<eta_m$ denote the ordered distinct elements of $A$, where $3leq mleq 9$ and
$$
begin{cases}
eta_1equiv mu_1-mu_1'-c-d\
eta_mequiv mu_1+a+b-mu_1'\
end{cases}
$$
Under assumption (2), $eta_1=-eta_m$, that is
$$
(diamond)hspace{1cm}mu_1-mu_1'=frac{c+d-a-b}{2}
$$

3) The candidates for $eta_2$ are
$$
begin{cases}
mu_1-mu_1'-c\
mu_1+a-mu_1'-c-d\
end{cases}
$$
and the candidates for $eta_{m-1}$ are
$$
begin{cases}
mu_1+a-mu_1'\
mu_1+a+b-mu_1'-c
end{cases}
$$
We know that
$$
{small
begin{aligned}
eta_1<min{mu_1-mu_1'-c,
& mu_1+a-mu_1'-c-d} leq max{mu_1-mu_1'-c,
mu_1+a-mu_1'-c-d} \
& leq
min{mu_1+a-mu_1',mu_1+a+b-mu_1'-c} leq max{mu_1+a-mu_1',mu_1+a+b-mu_1'-c}<eta_m
end{aligned}}
$$
We can show that, under assumption (2) and using $(diamond)$, this implies
$$
min{a,d}=min{b,c}
$$

4) Under assumption (1), $min{a,d}=min{b,c}$ reduces to two cases
$$
text{case i): } b=d<a,d<c
$$
and
$$
text{case ii): } a=c<b,c<d
$$

5)
We find $eta_1,...,eta_m$ under case i) and show that assumption (2) implies $mu_1=mu_1'$ and $a=c$.

6)
We find $eta_1,...,eta_m$ under case ii) and show that assumption (2) implies $mu_1=mu_1'$ and $b=d$.