Consider all $30$-dimensional vectors $v$ such that $v_i \in \{-1,1\}$. It seems that none of them has the property that $(v_1, \dots, v_{30})$ is orthogonal to $(v_{30}, v_1, \dots, v_{29})$.

Without just exhaustively enumerating all $2^{30}$ such vectors, how can one prove this?


This is true not just for $30$ but for all $n$ of the form $4k+2$. (Also all odd $n$, but only trivially)

Assume there exists such a vector such that $(v_1, \cdots, v_{4k+2}) \cdot (v_{4k+2}, v_1, \cdots, v_{4k+1}) = 0$ (s.t. they are othogonal). This can be written as the sum of $4k+2$ terms $$v_1v_{4k+2} + v_2 v_1 + \cdots + v_{4k+2} v_{4k+1}= 0 $$ since each $v_i \in \{1, -1\}$, each term is $\pm 1$. So $2k+1$ terms must be $1$ and $2k+1$ terms must be $-1$.

Now we count the $-1$ terms in the sum $v_1v_{4k+2} + v_2 v_1 + \cdots + v_{4k+2} v_{4k+1}$. If all $v_i$ are $1$, then there are zero terms that have value $-1$. Now consider an sum $v_1v_{4k+2} + v_2 v_1 + \cdots + v_{4k+2} v_{4k+1}$ with arbitrary $v_i$. Flipping the value of a $v_j$ for some $j \in \{1, \cdots, 4k+2\}$ will change the value of exactly two terms. Depending on the previous signs of these terms, the number of negative terms will be changed by $+2, 0$ or $-2$. So the number of negative terms in the sum is always even! This means that there can never be exactly $2k+1$ negative terms in the sum, so the dot product can never be $0$.

Set $k=7$ and done.


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