The fair-discrepancy propagation conjecture for fully balanced matrices

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Let A∈Mn×m(R)A\in\mathbb{M}_{n\times m}(\mathbb{R}) be a fully balanced matrix. For each row ii, let

Mi=1m∑j=1maijM_i=\frac{1}{m}\sum_{j=1}^{m}a_{ij}

be its average discrepancy. If, for a fixed 1≤i≤n1\leq i\leq n, one has

∣Mi−aij∣<ϵ|M_i-a_{ij}|<\epsilon

for the relevant entries jj, then ∣Mi−aij∣<ϵ|M_i-a_{ij}|<\epsilon for all 1≤i≤n1\leq i\leq n.

Fair-discrepancy propagation conjecture. If one fixed row has discrepancy less than ϵ\epsilon, then every row has discrepancy less than ϵ\epsilon.

The claim concerns propagation of a fair discrepancy from one row to all rows of a fully balanced matrix. The supplied text gives no resolution or additional context.

References

Primary source

Theophilus Agama and Gael Kibiti, “Balanced matrices”, arXiv:1810.07542 (2026).

Progress summary

Refreshed
Claimed progress

A reader-submitted example appears to disprove the conjecture, but it has not been independently checked, so the general question remains open.

The statement is recorded as Conjecture 6.6 in Balanced matrices: fair discrepancy in one row of a fully balanced matrix should propagate to every row. No published proof or disproof was found.

Known results

  • For positive fully balanced matrices of size 2×22\times2, fair discrepancy along rows is equivalent to fair discrepancy along columns (Theorem 6.4).
  • In the same 2×22\times2 setting, fair discrepancy in one row implies fair discrepancy in all rows (Proposition 6.5).

Community submission (unverified)

The August 22, 2026 submission argues that A=(555157751)A=\begin{pmatrix}5&5&5\\1&5&7\\7&5&1\end{pmatrix} has equal row and column squared energies, while its first row has zero discrepancy and its second row violates the condition for ε=1\varepsilon=1; scaling is claimed to extend this to every ε>0\varepsilon>0.

Current status (as of September 2026): The positive 2×22\times2 case is settled, while the general conjecture remains open and the submitted counterexample is unverified.

Sources

Solutions 1

CounterexampleThis solution needs a summarySee full solutionHide full solution

Counterexample to discrepancy propagation, even for strictly positive, exactly balanced matrices.

For a matrix A=(aij)∈Rn×mA=(a_{ij})\in\mathbb{R}^{n\times m}, write

Ri=∑j=1maij2,Cj=∑i=1naij2,Mi=1m∑j=1maij.R_i=\sum_{j=1}^{m}a_{ij}^{2},\qquad C_j=\sum_{i=1}^{n}a_{ij}^{2},\qquad M_i=\frac{1}{m}\sum_{j=1}^{m}a_{ij}.

The conjecture asserts that, for a fully balanced matrix and a fixed ε>0\varepsilon>0, the inequalities

∣Mr−arj∣<ε(1≤j≤m)\lvert M_r-a_{rj}\rvert<\varepsilon \qquad (1\le j\le m)

for one row rr imply the corresponding inequalities in every row. Consider the strictly positive integer matrix

A=(555157751).A= \begin{pmatrix} 5&5&5\\ 1&5&7\\ 7&5&1 \end{pmatrix}.

Its row energies are

R1=52+52+52=75,R2=12+52+72=75,R3=72+52+12=75,\begin{aligned} R_1&=5^2+5^2+5^2=75,\\ R_2&=1^2+5^2+7^2=75,\\ R_3&=7^2+5^2+1^2=75, \end{aligned}

and its column energies are

C1=52+12+72=75,C2=52+52+52=75,C3=52+72+12=75.\begin{aligned} C_1&=5^2+1^2+7^2=75,\\ C_2&=5^2+5^2+5^2=75,\\ C_3&=5^2+7^2+1^2=75. \end{aligned}

Consequently, AA is simultaneously horizontally and vertically balanced with exact equality, which is stronger than the approximate balance required in the original definition. Set ε=1\varepsilon=1. The first row has mean

M1=5+5+53=5,M_1=\frac{5+5+5}{3}=5,

and therefore

∣M1−a1j∣=0<1(j=1,2,3).\lvert M_1-a_{1j}\rvert=0<1 \qquad (j=1,2,3).

However, the second row has mean

M2=1+5+73=133,M_2=\frac{1+5+7}{3}=\frac{13}{3},

so its first entry satisfies

∣M2−a21∣=∣133−1∣=103>1.\lvert M_2-a_{21}\rvert =\left\lvert\frac{13}{3}-1\right\rvert =\frac{10}{3}>1.

Thus the first row has fair discrepancy while the second does not, disproving the conjecture.

In fact, the obstruction persists for every prescribed tolerance: for any ε>0\varepsilon>0, choose t>3ε/10t>3\varepsilon/10 and replace AA by tAtA. Every row and column then has squared energy 75t275t^2; the first-row deviations remain zero, whereas

∣M2(tA)−(tA)21∣=10t3>ε.\left\lvert M_2(tA)-(tA)_{21}\right\rvert =\frac{10t}{3}>\varepsilon.

If desired, taking t≥1t\ge1 also preserves the stronger entrywise condition (tA)ij≥1(tA)_{ij}\ge1. Hence neither exact energy balance nor strict positivity restores discrepancy propagation.

Source: T. Agama and G. Kibiti, Balanced matrices, arXiv:1810.07542v3, Definition 3.1, Conjecture 6.6, and Remark 6.7.