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How do I construct two vectors $a,b\in \mathbb{R}^{n}$, $a=(a_1,a_2,\ldots, a_n)^T$ and $b=(b_1,b_2,\ldots, b_n)^T$ which satisfy in the following conditions ‎\begin{align} & a_ib_i\geq 1,a_ib_j<1, i\neq j, \quad i,j=1,2,\ldots, n,\\ & \hspace{40 mm} \text{or}\\ & a_ib_i> 1,a_ib_j\leq1, i\neq j, \quad i,j=1,2,\ldots, n. ‎\end{align}

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    $\begingroup$ So, you want a rank 1 matrix whose diagonal entries are at least 1, off-diagonal entries less than 1. $\endgroup$ Commented Jan 3, 2019 at 14:28

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Using the rewording of the problem by Gerry Myerson, the analysis below shows it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, and let the entries of the first vector be denoted $a,b,c$, with those in the second vector $d,e,f$, so that outer-product matrix is given by $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

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