The problem asks whether the existence of mutually orthogonal non-zero vectors (\mathbf{1}, \mathbf{v_1}, \ldots, \mathbf{v_{n-1}} \in {-1,0,1}^n) implies (n \in {1,2} \cup {4k : k \in \mathbb{N}}).

Key steps:

1. Orthogonality and Support Analysis: The all-ones vector (\mathbf{1}) is orthogonal to each (\mathbf{v_i}), so (\sum v_i(k) = 0). This implies each (\mathbf{v_i}) h...

Yes. If there exist n mutually orthogonal nonzero vectors 1, v2, …, vn ∈ {−1,0,1}^n with the first one equal to the all-ones vector 1, then necessarily n ∈ {1,2} ∪ {4k : k ∈ N}.

Here is a self-contained argument that yields the necessary condition.

Set-up and notation

• Let the n vectors be the rows of an n×n integer matrix M whose entries are in {−1,0,1}, with the first row equal to 1 = (1, …, ...

Short answer
Nobody presently knows. At the moment every construction that is known works only for
n = 1 , 2 or for n divisible by 4,
and every systematic search that has been carried out has failed to find an example for any other value of n; but no proof ruling the other values out is known.

What is known so far

1. Relation with Hadamard and weighing matrices.
   • If all ...

Here is the state‐of‐the‐art on this problem, together with the short “determinant + mod 2’’ proof that no new orders appear beyond the usual Hadamard‐orders.

1. Reformulation as a weighing‐matrix problem.
   Stack your (n) vectors as the rows of an (n\times n) matrix [ W ;=; \begin{pmatrix} 1 & 1 & \cdots & 1 \[6pt] v_1 & & ...

This is an excellent question that sits at the intersection of linear algebra, combinatorics, and number theory. The answer to the question is yes, and the reasoning breaks down into analyzing the properties of the integer .

Let the set of vectors be , where (the all-ones vector) and $\mathbf{u_i} = \math...

<think> I need to determine if it's true that if you have a set of mutually orthogonal non-zero vectors in \(\{-1, 0, 1\}^n\) including the all-ones vector, then n must be 1, 2, or a multiple of 4. From the notes, this is known for \(\{-1, 1\}\) without zeros, which is the Hadamard matrix case. With zeros allowed, it might be different, but the user has checked for small n and found no such vector...

Summary of what is presently known

Write
v0 = 1 = (1,1,… ,1)
v1 ,…, v n–1 ∈ { –1 ,0 ,1 }^n, n ≥ 1,

and assume that the n vectors v0 ,…, v n–1 are pairwise orthogonal in the usual Euclidean inner product and that every vi is non-zero.
Put M =(v0 ,…, v n–1 ) (so the vi are the rows of an n×n integral matrix M). The question is for which n such a matrix can exist.

1. T...

Analyzing Orthogonal Vectors with Entries from {-1,0,1}

To determine whether is a necessary condition for the existence of mutually orthogonal non-zero vectors , I'll analyze the constraints imposed by orthogonality.

Initial Observations

First, let's understand ...

Looking at this problem, I need to analyze when we can have mutually orthogonal non-zero vectors in where one of them is the all-ones vector.

Let me denote these vectors as .

Key Observations:

1. Since , we have for each .

2. Since the vectors ar...