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Question. Let $A=(a_{ij})\in\mathbb R^{(2n+1)\times (2n+1)}$, while $a_{ii}=0$, for all $i=1,\ldots,2n+1$, and $a_{ij}\in\{-1,1\}$, if $i\ne j$. If $$ \sum_{j=1}^{2n+1}a_{ij}=0, \quad \text{for all $i=1,\ldots,2n+1$}, $$ show that $\mathrm{Rank}(A)=2n$.

Clearly, $A\boldsymbol{u}=0$, where $\boldsymbol{u}=(1,1,\ldots,1)$, and hence $\mathrm{Rank}(A)\le 2n$, and $\mathrm{Ker}A\ne\{0\}$. Another piece of information which could be useful is that the kernel of $A$ is spanned by eigenvectors with rational (or even integer) elements.

Yiorgos S. Smyrlis
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  • I don't know whether it would work, but you may try by contradiction? Like assume that $v_1, v_2 \in Ker A$ are independent, then show that $v_1 = \lambda v_2$ by doing some trickery with $A(v_1 + v_2)= 0$? – G. Gare Dec 05 '22 at 08:11

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Let $M$ be the leading principal $2n\times2n$ submatrix of $A$. Then $M\equiv I_{2n}-\mathbf1\mathbf1^T$ modulo $2$ and in turn, $\det(M)\equiv1-\mathbf1^T\mathbf1\equiv1$ modulo $2$. Hence $\det(M)$ is an odd integer and the rank of $A$ is at least $2n$.

user1551
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    It's not as direct, but equivalently, one may prove that the number of derangements of ${1,\dots,2n}$ is odd. (then write the determinant as a sum over permutations, remove the $0$ terms and you are left with a sum of an odd number of terms $\pm1$). – Jean-Claude Arbaut Dec 05 '22 at 08:26
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    another option, directly over $\mathbb R$, confirm $\det\big(B\big)\neq 0$, when $B:=A+\mathbf {11}^T$ since determinant is a product of 1s on the diagonal plus a sum of products consisting solely of even numbers hence determinant is odd $\neq 0$ – user8675309 Dec 05 '22 at 17:32
  • @user1551 I am afraid, I do not understand something in your proof. If $M=(a_{ij}){i,j=1}^{2n}$, then the diagonal elements of $M$ are still equal to zero. Meanwhile, the diagonal elements of the matrix $I{2n}-\boldsymbol{11}^T,$ are all equal to $1-2n$, and hence equal to $1$ modulo $2$. – Yiorgos S. Smyrlis Dec 06 '22 at 09:36
  • @YiorgosS.Smyrlis No, $\mathbf1\mathbf1^T$ is the all-one matrix. Hence $I_{2n}-\mathbf1\mathbf1^T$ has a zero diagonal. – user1551 Dec 06 '22 at 09:53
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First of all, this question originates from the following old Soviet Olympiad problem (see also here):

Given 13 gears each weighing an integral number of grams. It is known that any twelve of them may be placed on a pan balance, six on each pan, in such a way that the scale will be in equilibrium. Prove that all these gears must be of equal weight.

It turns out that the requirement the the gears should be integral numbers of grams can be relaxed. It is still true even if the gears are real numbers of grams. Also the number 13 could be replaced by any odd positive integer.

If $\boldsymbol v=(v_1,\ldots,v_{13})$ a set of gears satisfying the requirements of the problem, then the vector $\boldsymbol v\,$ belongs to the kernel of a $13\times13$ matrix $A$ of the form of the OP. Clearly, $(1,1,\ldots,1)\in \mathrm{Ker}\,A$. It suffices to show that $\mathrm{Rank}(A)=12$.

Assume now we have proved the case in which the gears are integers. (See here.) We next observe that the Kernel of $A$ is spanned by vectors with rational elements (i.e., using Gauss-Jordan). In fact, it is spanned with vectors with integral elements. But the only such vectors with integer elements are the multiples of $(1,1,\ldots,1)$.

Yiorgos S. Smyrlis
  • 86,581