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Let A be a $n \times n$ matrix such that:

1) all the diagonal elements are $0$,

2) all other elements are either $1$ or $-1$

3) number of $1$'s in a row are equal to number of $-1$'s in the row, i.e. the row-sum is $0$ for all rows.

It is obvious that $n$ is odd here. I have done some experiments and found that the rank of such an $A$ is always $n-1$. Can someone help me with a proof? I can only see that

\begin{bmatrix} 1 \\ 1\\ \vdots\\ 1\\ 1\\ \end{bmatrix}

is an eigen vector with eigen value =0.

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    as an starting point: try to sum two columns such that one row is a zero. – 2017-01-09
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    I have checked your claim for $A = \begin{bmatrix} 0 & x & -x \\ y & 0 & -y \\ z & -z & 0 \end{bmatrix}$, it is fabulous! – 2017-01-09
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    @Math-fun, I did not exactly understand what you mean. We don't have the exact values of the entries just the information that sum of rows is 0. – 2017-01-09
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    @Math-fun Summing up all the columns will give me a vector with all elements equal to 0, if this is what you meant. – 2017-01-09
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    very good! now, what do you conclude about linear relationships between columns? what does it imply for rank? – 2017-01-09
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    @Math-fun that they are Linearly dependent and that the rank $\leq$ n-1. – 2017-01-09
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    very good :-) Can now the rank be even smaller than $n-1$? – 2017-01-09

2 Answers 2

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Look at the leading principal $(n-1)\times(n-1)$ submatrix $A_{n-1}$. In modulo-2 arithmetic, $A_{n-1}=ee^T-I_{n-1}$, where $e^T=(1,\ldots,1)$. Its eigenvalues are $n-2$ and $-1$ modulo $2$, which are nonzero because $n$ is odd. So $\det A_{n-1}\ne0$ in modulo-2 arithmetic and $\det A_{n-1}\ne0$ over $\mathbb R$ too. Hence $A_{n-1}$ is nonsingular and $A$ has rank $n-1$.

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    Very witty, as usual – 2017-01-09
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    Basically you are making all the -1's as 1's. – 2017-01-09
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    I think your answer requires an explanation. How can we say that $det((A)mod(2)) = (det(A))mod(2)$? – 2017-01-09
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    @AbhayGupta Let $A_{n-1}=2B+R$, where $B$ is an integer matrix and all entries of $R$ belong to $\{0,1\}$. What I am saying is that $\det A_{n-1}\equiv\det R\mod2$ and therefore $\det A_{n-1}=0$ if and only if $\det R\equiv0\mod2$. You may prove this directly using the formula $\det A=\sum_{\sigma}\operatorname{sign}(\sigma)\prod_ia_{i\sigma(i)}$, or prove it by mathematical induction, with determinants calculated by Laplace expansion. Basically this is just consequence of the identities $(2p+r)(2q+s)\equiv rs\mod2$ and $(2p+r)+(2q+s)\equiv r+s\mod2$. – 2017-01-09
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    @user1551 I faced the same question again today and I forgot why "Its eigenvalues are $n-2$ and $-1$ mod $2$"? – 2017-08-16
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    $n-2$ is due to the eigen vector with all $1's$. Is $-1$ due to eigen vector with equal $1's$ and $-1's$. If yes aren't all these vectors the same in mod $2$? – 2017-08-16
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    @AbhayGupta $ee^T$ is an $(n-1)\times(n-1)$ rank-1 matrix. So, it has $n-2$ zero eigenvalues. The remaining eigenvalue is $n-1$ because $e$ is an eigenvector. Hence the spectrum of $A_{n-1}=ee^T-I_{n-1}$ is obtained by translating the spectrum of $ee^T$ to the left by 1. – 2017-08-16
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    @AbhayGupta That all eigenvalues of $A_{n-1}$ are equal to $1$ doesn't make $e=(1,1,\ldots,1)^T$ the only eigenvector of $A_{n-1}$. E.g. when $n=5$, the 4x4 matrix $A_{n-1}$ has three eigenvectors $(1,1,1,1)^T,(1,0,1,0)^T$ and $(1,0,0,1)^T$. Note that $A_{n-1}$ is **not** diagonalisable (the only diagonalisable matrix with all eigenvalues equal to 1 is the identity matrix, which $A_{n-1}$ is not), but it has a generalised eigenvector $(0,0,0,1)^T$. If you put these four vectors to form a transformation $P$, then $P^{-1}AP$, up to permutation, is a Jordan form with exactly one 2x2 Jordan block. – 2017-08-16
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I'll outline a proof ...

Let $A$ be an $n$ x $n$ matrix of the specified type.

As noted, $n$ must be odd.

Let $B$ be the $(n-1)$ x $(n-1)$ upper left-hand submatrix of $A$.

The determinant of $B$ is the sum of the products of the generalized diagonals.

Of the $(n-1)!$ such products, the nonzero products are odd (equal to either $+1$ or $-1$).

The number of diagonals whose product is nonzero is the number of derangements of $\{1,...,n-1\}.$

Since $n-1$ is even, the number of derangements is odd.

Thus, $\text{det}(B)$ is the sum of an odd number of odd numbers, so is odd, and hence is nonzero.

It follows that $\text{rank}(A) \ge n-1$.

Since the columns of $A$ sum to zero, $\text{rank}(A) \le n-1$.

Therefore $\text{rank}(A) = n-1$.