$A$ does not reverse the sign of any vector iff $A^{T}$ does not. I want to prove the following: $A$ does not reverse the sign of any vector (that is, if $x \neq 0$ and $y=Ax$, then for some subscript $i$, $x_{i}y_{i}>0$) if and only if $A^{T}$ does not reverse the sign of any vector. 
Let us assume that $A$ does not reverse the sign of any vector. $x \neq 0$, $y=Ax$ and $i$ be the index such that $x_{i}y_{i}>0$. Let $z= A^{T}x$. I need to choose an index $k$ such that $x_{k}z_{k}>0$. But I don't know how to choose it! Please help me!
 A: We will show that following are equivalent for $A \in \operatorname{Mat}_n(\mathbf R)$:

(1) $A$ does not reserve the sign of any vector.
(2) All real eigenvalues of $A$ and all its principal submatrices are positive.
(3) All principal minors of $A$ are positive.

Proof. (1) $\to$ (2): As any prinicipal submatrix of $A$ does not reverse the sign of any vector, it suffices to show, that any real eigenvalue of $A$ is positive. Suppose $Ax = \lambda x$ for $x \in \mathbf R^n \setminus \{0\}$. Then for some $i$, by (1) 
$$ 0 < x_i(Ax)_i = \lambda x_i^2 \iff \lambda > 0 $$
(2) $\to$ (3): Let $B$ be a principal submatrix of $A$. As the determinant is the product of the eigenvalues, and the real eigenvalues are positive, and for a complex eigenvalue $\lambda$ we have $\bar\lambda$ is also an eigenvalue, we have 
$$ \det B = \prod_{i : \lambda_i \in \mathbf R} \lambda_i \cdot \prod \lambda\bar\lambda = \prod_i \lambda_i \cdot \prod |\lambda|^2 > 0 $$
(3) $\to$ (1): We will use induction on $n$. For $n=1$, we have nothing to prove, as $A$ is the only principal minor of $A$, and $A$ does not reverse signs iff $A > 0$. Now let $n > 1$. Suppose $A$ reverses the sign of $x \in \mathbf R^n$, if $x_i = 0$, then the principal submatrix $(a_{jk})_{j,k\ne i}$ reverses the sign of $(x_j)_{j\ne i}$, which is wrong by induction hypothesis. Hence $x_i \ne 0$ for all $i$. We may define $d_i := \frac{(Ax)_i}{x_i}$ with $d_i \le 0$, as $A$ reverses $x$'s sign. Let $D := \operatorname{diag}(-d_1, \ldots, -d_n)$. Then 
$$ \det(A + D) = \sum_{I \subseteq \{1, \ldots, n\}} \det(D_{II})\det M_{II} > 0$$
But $(Dx)_i = -d_i x_i = -(Ax)_i$, that is $(A+D)x = 0$. Contradiction. So (1) is true. $\Box$
As $A$ and $A^t$ have the same principal minors, the statement follows from $(1) \iff (3)$.
