TITLE: Real orthogonal and sign
QUESTION [6 upvotes]: I came across the following conjecture, reading a recent paper in the Monthly, an orthogonal matrix of order $n\neq 0 \pmod 4$ has a nonnegative  (up to a scalar) row vector.
It should be straight  in dimensions $2$ and $3$.

REPLY [12 votes]: Here's a simple argument that this is false for $n > 2$.
In dimension $n$ there are $2^{n}$ orthants, $2^{n-1}$ if one considers them modulo sign. A pair of antipodal orthants means a pair of orthants whose sign patterns are exactly opposite (e.g. the positive orthant union the negative orthant).
The key point is that an orthonormal frame contains $n$ vectors, no two of which lie in the same pair of antipodal orthants. If two nonzero vectors lie in the interior of the same orthant, then their inner product is positive because it is a sum of nonnegative numbers not all of which are zero. (If they lie on the boundary of the same orthant, a small rotation moves them to different orthants).
The original "conjecture" supposes that necessarily one of the vectors of an orthonormal frame lies in a particular orthant. However, since $2^{n-1} > n$ if $n > 2$, necessarily some pair of antipodal orthants contains no vector of the frame (pigeonhole principle). Since permutations of the coordinates are orthogonal transformations permuting the orthants (and preserving antipodal orthants), given a particular orthant there is always an orthonormal frame containing no vector of the orthant.
(Note that the preceding is a bit vague in that it is not specified whether orthants are open or closed; since the goal is to build a counterexample, this does not matter much.)