Fact. Let $u\in\mathbb{R}^n$, $u\neq 0$. Suppose that $u$ is orthogonal to precisely $2^{n-1}$ sign-vectors (i.e., vectors $\varepsilon\in\mathbb{R}^n$ such that $\varepsilon_i=\pm 1$ for each $1\leq i\leq n$.) Then $u$ has precisely two non-zero coordinates which have the same absolute value.

Does anyone have a reference for a proof of this fact, or a proof without a reference? I have a proof, but it is a bit long and somewhat tricky.

  • $\begingroup$ If the formal proof is long and tricky, it might be easier to just make a heuristic argument instead of proving it formally. I've found it's a lot simpler to argue heuristicly when dealing with vector spaces as formal proofs can become very pendantic very quickly. $\endgroup$
    – Patty
    Commented Jun 18, 2017 at 22:41
  • 1
    $\begingroup$ Have you heard of Hadamard matrices (en.wikipedia.org/wiki/Hadamard_matrix)? $\endgroup$
    – Jean Marie
    Commented Jun 18, 2017 at 22:42
  • $\begingroup$ @JeanMarie. Yes, I have. Can you apply them for $n=13$? $\endgroup$ Commented Jun 18, 2017 at 22:42
  • $\begingroup$ @Patty - I'd love to hear a heuristic argument! $\endgroup$ Commented Jun 18, 2017 at 22:43
  • 1
    $\begingroup$ @JeanMarie - For every $n\geq 2$, the vector $(1,1,0,\dots,0)$ satisfies the hypothesis. The problem is to show that this is essentially the only one, up to scaling and a cube-symmetry. $\endgroup$ Commented Jun 19, 2017 at 7:37

4 Answers 4


I find it simpler to deal with sets than with orthogonal vectors, so I am going to translate the problem:

For $v=(v_1,\dots,v_n)\in \Bbb{R}^n$ let $E_v$ be the sets of subsets $X$ of $\Bbb{N}_n:=\{1,\dots,n\}$ such that $$ (1)\quad\quad\sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i. $$ Clearly there is a bijection between the sets $X$ in $E_v$ and the sign vectors which are orthogonal to $v$, assigning to every set $X$ the sign vector $S_X:=2e_X-\Bbb{I}$, where $\Bbb{I}$ is the vector with all entries equal to $1$ and $(e_X)_i=1$ if $i\in X$ and $0$ otherwise. If $X\in E_v$, then we have $$ v\cdot e_X=\sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i=\frac 12 v\cdot\Bbb{I}, $$ and so $v\cdot S_X=0$. On the other hand $v\cdot S_X=0$ implies $X\in E_v$.

For any $i\in \Bbb{N}_n$ with $v_i\ne 0$, we can define an injective map $R_i$ from $E_v$ into $P(\Bbb{N}_n)\setminus E_v$ (the complement of $E_v$ in the powerset of $\Bbb{N}_n$) defining $R_i(X)=X\triangle\{i\}$, i.e., $R_i(X)=X\cup\{i\}$ if $i\notin X$ and $R_i(X)=X\setminus\{i\}$ if $i\in X$. Clearly, if (1) is valid for $X$, then it is not valid for $R_i(X)$.

Now, if $\#(E_v)=2^{n-1}$, then $R_i$ is a bijection for every $i$ with $v_i\ne 0$, and it is clearly its own inverse.

Assume by contradiction that there exist three non-zero entries $v_{i_1},v_{i_2},v_{i_3}$, and set $X=\Bbb{N}_n\setminus\{i_1,i_2,i_3\}$.

If $X\in E_v$, then $R_{i_1}(X)\notin E_v$ and so $R_{i_2}(R_{i_1}(X))\in E_v$, similarly $R_{i_3}(R_{i_1}(X))\in E_v$, hence $$ v_2+v_1+\sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i=v_3+v_1+\sum_{i\in X}v_i, $$ and so $v_2=v_3$. Simlarly one proves $v_1=v_2$. But $X\in E_v$, so $$ \sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i=v_2+v_1+\sum_{i\in X}v_i, $$ and so $v_1=v_2=0$, a contradiction.

If $X\notin E_v$, then $R_{i_1}(X),R_{i_2}(X),R_{i_3}(X)\in E_v$, hence $$ v_1+\sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i=v_2+\sum_{i\in X}v_i=v_3+\sum_{i\in X}v_i, $$ and so $v_1=v_2=v_3$. But then $R_{i_2}(R_{i_1}(X))\notin E_v$ which implies that $R_{i_3}(R_{i_2}(R_{i_1}(X)))\in E_v$ and so $$ v_1+v_2+v_3+\sum_{i\in X}v_i=\frac 12 \sum_{i\in \Bbb{N}_n}v_i=v_1+\sum_{i\in X}v_i, $$ which leads to $v_2=v_3=0$, also a contradiction.

Hence there are at most two nonzero entries, and clearly they are either equal or they are opposite, and it is also clear that no vector with only one nonzero entry satisfies (1).

  • $\begingroup$ Thank you! This is very nice. $\endgroup$ Commented Jun 22, 2017 at 14:01

The following more general version of your question was originally raised by Littlewood and Offord in their study of the roots of random polynomials.

Let $\{a_1, a_2, \dots, a_k\}$ be $k$ (not necessarily distinct) real numbers of magnitude at least $1$. What is the maximum number of expressions of the form $$\pm a_1 \pm a_2 \dots \pm a_k$$ that can lie in an interval of length less than $2$?

They proved a bound that was within a $\log k$ factor of optimal. Shortly afterwards, Erdős showed that the correct bound is $\binom{k}{\lfloor k/2 \rfloor}$. Taking his result, rescaling it, and padding the sequence out with zeros gives the following:

Let $\{a_1, a_2, \dots, a_n\}$ be real numbers, $k$ of which are nonzero. Then for any $c$, the number of solutions to $$\pm a_1 \pm a_2 \dots \pm a_n = c$$ is at most $\binom{k}{\lfloor k/2 \rfloor} 2^{n-k}$. Equality holds iff the nonzero terms are all equal in magnitude.

Erdős's argument is actually a close cousin of San's: WLOG we can assume that all of the $a_i$ are non-negative. Then associate each sign sequence with the set corresponding to the indices of nonzero terms with positive signs. Changing negative signs to positive signs increases the sum, so the sets corresponding to equal sums form an antichain (no set is a subset of another). The bound follows from Sperner's Theorem.

In terms of your problem, you're stating that the ratio $\binom{k}{\lfloor k/2 \rfloor}/2^k$ is equal to $\frac{1}{2}$, which only happens for $k=1$ and $k=2$, and $k=1$ doesn't give $c=0$.

  • $\begingroup$ Very nice! How do you deduce from the equality case in Sperner's theorem that all non-zero terms are equal in magnitude? $\endgroup$ Commented Jun 23, 2017 at 23:07
  • $\begingroup$ Equality holds in Sperner's theorem only when the set is a slice of the hypercube (either all sets of size $\frac{k}{2}$, if $k$ is even, or either all sets of size $\frac{k+1}{2}$ or all sets of size $\frac{k-1}{2}$, if $k$ is odd). If equality holds in Littlewood-Offord, then all subsets of the corresponding size must have the same sum, which implies that all of the summands must be equal. $\endgroup$ Commented Jun 23, 2017 at 23:26

This is just a translation of san's beautiful argument to a geometric language, with a few short-cuts.

By symmetry we may assume that $$u_1\geq u_2\geq\cdots \geq u_n\geq 0\quad\quad\quad\quad\quad(1)$$ If $n=2$ then the only vectors orthogonal to sign-vectors are proportional to sign-vectors themselves, so the assertion is clear. Assume henceforth that $n\geq 3$. Put $$E_u=\{\varepsilon\in\{-1,1\}^n: \langle\varepsilon,u\rangle = 0\}$$ Consider the $i$'th coordinate "sign-flip", i.e., the cube symmetry $R_i$ defined by multiplying the $i$'th coordinate of a vector by $-1$: $$R_i(x_1,\cdots,x_i,\cdots x_n)=(x_1,\cdots,-x_i,\cdots x_n)$$

There are two crucial properties of $R_i$ in connection with our problem, which are easy to check:

  1. For each $i$ such that $u_i\neq 0$, $R_i$ is an injection of $E_u$ into its complement $(E_u)^c$ in $\{-1,1\}^n$, and if $|E_u|=2^{n-1}$ it is a bijection.

  2. $R_i^2=Id$

Now assume that $u_3>0$. Put $\varepsilon=(-1,-1,-1,+1,\dots,+1)$.

Case A: $\varepsilon\in E_u$.

$R_2\varepsilon$ does not belong to $E_u$, and since $R_1$ is a bijection such that $R_1^2=Id$, the sign-vector $R_1(R_2\varepsilon)$ belongs to $E_u$. That is: $$u_1+u_2-u_3=-\sum_{i=4}^nu_i$$ By (1), the r.h.s is not positive and the l.h.s is not negative, so they both must be zero, implying that $u_1+u_2=u_3$, which (by (1) again) is impossible unless $u_1=u_2=u_3=0$, contradiction.

Case B: $\varepsilon\notin E_u$

Flipping once with $R_1$ takes $\varepsilon$ to $E_u$. Flipping again with $R_2$ takes $R_1\varepsilon$ out of $E_u$, but flipping a third time with $R_3$ takes it back, so $R_1(R_2(R_3\varepsilon))\in E_u$. Hence: $$u_1+u_2+u_3=-\sum_{i=4}^nu_i$$ and we arrive at a contradiction as before.

It follows that $u_3=0$, and so there are only two non-zero coordinates - and so $u_1=u_2$ as in the two-dimensional case above.

  • $\begingroup$ Nice translation and nice shortcuts. Maybe $E_v$ should be $E_u$. $\endgroup$
    – san
    Commented Jun 22, 2017 at 14:57
  • $\begingroup$ Thank you. Of course, $E_u$. $\endgroup$ Commented Jun 22, 2017 at 15:21

Partial solution: I have made this community wiki so that anyone can finish the answer if they know how (otherwise, feel free to just create a new answer with your full solution) Let $E$ be the set of sign vectors which are orthogonal to $v$. If, for all $e\in E$, we have $e_1=e_2$, then this set must be exactly $$ E=\{(1,1,\pm1,\dots,\pm1)\}\cup\{(-1,-1,\mp1,\dots,\mp1)\} $$ which can be seen by the involution of multiplication by $-1$. Now $v$ is perpendicular to $(1,1,1,\dots,1)$ and $(1,1,-1,\dots,-1)$ so it is perpendicular to the sum, $(2,2,0,\dots,0)$. Thus $v_1=-v_2$. If both $v_1=v_2=0$, then $E$ would not be the set described above, so we must have $|v_1|=|v_2|\neq0$. Given the description of $E$ above it is now easy to see that the last $n-2$ coordinates of $v$ are $0$.

In fact, the above logic works if $e_i=e_j$ for all $e\in E$ and any indices $i,j$, and similarly if $e_i=-e_j$ for any $i,j$.

Now it suffices to show that no such $E$ can exist which doesn't meet those restrictions. I am thinking that we should use the fact that $E$ is closed under multiplication by $-1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .