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Consider the $3\times 3$ matrix $$ A\equiv \begin{pmatrix} \mu_1-\mu_1'& \mu_1-\mu_1'-c & \mu_1-\mu_1'-c-d\\ \mu_1+a-\mu_1'& \mu_1+a-\mu_1'-c & \mu_1+a-\mu_1'-c-d\\ \mu_1+a+b-\mu_1'& \mu_1+a+b-\mu_1'-c & \mu_1+a+b-\mu_1'-c-d\\ \end{pmatrix} $$ 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) $a\neq b,c\neq d$

(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. $$ A\equiv \begin{pmatrix} \mu_1-\mu_1'&<& \mu_1-\mu_1'-c &<& \mu_1-\mu_1'-c-d\\ \wedge && \wedge && \wedge\\ \mu_1+a-\mu_1'&<& \mu_1+a-\mu_1'-c &<& \mu_1+a-\mu_1'-c-d\\ \wedge && \wedge && \wedge\\ \mu_1+a+b-\mu_1'&<& \mu_1+a+b-\mu_1'-c &<& \mu_1+a+b-\mu_1'-c-d\\ \end{pmatrix} $$

2) Let $\eta_1<\eta_2<...<\eta_m$ denote the ordered distinct elements of $A$, where $3\leq m\leq 9$ and $$ \begin{cases} \eta_1\equiv \mu_1-\mu_1'-c-d\\ \eta_m\equiv \mu_1+a+b-\mu_1'\\ \end{cases} $$ 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 $b=d$.

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    $\begingroup$ Also, $A = (\mu_{1}-\mu_{1}^{\prime}) \begin{bmatrix} 1&1&1\\1&1&1\\1&1&1\end{bmatrix} + a\begin{bmatrix}0&0&0\\1&1&1\\1&1&1\end{bmatrix} +b\begin{bmatrix}0&0&0\\0&0&0\\1&1&1\end{bmatrix} -c\begin{bmatrix}0&1&1\\0&1&1\\0&1&1\end{bmatrix} -d\begin{bmatrix}0&0&1\\0&0&1\\0&0&1\end{bmatrix}$, which will be symmetric iff $a=c$ and $b=d$. $\endgroup$ – Morgan Rodgers Mar 21 at 18:18
  • $\begingroup$ @MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now? $\endgroup$ – STF Mar 21 at 18:21
  • $\begingroup$ @MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $\mu_1+c-\mu_1'=\mu_1-\mu_1'-c$ which is impossible given $c>0$. $\endgroup$ – STF Mar 21 at 18:35
  • $\begingroup$ I apologize, I meant to say $a=-c$, $b=-d$. $\endgroup$ – Morgan Rodgers Mar 21 at 19:01
  • $\begingroup$ Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric. $\endgroup$ – STF Mar 21 at 19:03

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