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 $3$.



Real orthogonal and sign
This is **false** already for $n=3$. A counterexample is given by the matrix 


$$A=\frac{1}{3}\begin{bmatrix}
2 & 2 & -1\\
2 & -1 & 2\\
-1& 2 & 2
\end{bmatrix}.
$$


How did I construct this matrix? Start with $(a,b,c)$ such that $ab+bc+ca=0$, where $a,b,c\ne 0$. Observe that $a,b,c$ cannot be of the same sign. Renormalize it to be of unit length, consider the corresponding (anti-)circulant matrix $A$ and VOILA! Solving the equation is easy: $\displaystyle c=-\frac{ab}{a+b}.$


More generally, for any $n\geq 3$, consider a real unit row-vector ${\bf a}=(a\_1,\ldots,a\_n)$ such that all entries are non-zero and ${\bf a}$ is orthogonal to its cyclic permutations, i.e. $a\_{i}\ne 0,\, \sum\_{i=1}^n a\_{i}^2=1$ and $$\displaystyle\sum\_{i=1}^n a\_{i}a\_{i+p}=0{\ \ }{\rm for}{\ }1\leq p\leq n-1,\tag {\*}$$ where $a\_{j+n}=a\_{j}.\ $ Then the matrix $A$ whose rows are successive cyclic permutations of the row vector ${\bf a}$ is orthogonal with non-zero entries, but has no row with entries of the same sign. To construct such a vector, one only needs to solve a system of $\displaystyle m=\lfloor\frac{n}{2}\rfloor$ homogeneous quadratic equations (\*) with $1\leq p\leq m$ in non-zero reals and renormalize the solution to be a unit vector. This is straightforward, but the hour is late, so let me leave it as an exercise to the interested readers.


