# Rewrite a condition on a $3\times3$ matrix

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 $$\text{case i): } b=d and $$\text{case ii): } a=c

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 $$b=d$$.

• 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 $b=d$. – Morgan Rodgers Mar 21 at 18:18
• @MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now? – STF Mar 21 at 18:21
• @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 $c>0$. – STF Mar 21 at 18:35
• I apologize, I meant to say $a=-c$, $b=-d$. – Morgan Rodgers Mar 21 at 19:01
• 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. – STF Mar 21 at 19:03