There exist an element $\textbf {x}\in \mathbb {F}^n$ with all $x_i\neq0$ such that $A\textbf{x}$ has all non-zero coordinates. 
Let $\mathbb{F}$ be an infinite field  (possibly of characteristic $0$) and $A$ be an $m\times n$ matrix over $\mathbb{F}$ having all non-zero rows. Want to show there exist an element $\textbf {x}=(x_1,\ldots,x_n) \in \mathbb {F}^n$ with all $x_i\neq0$ such that $A\textbf{x}$ as an element in $\mathbb{F}^m$ has all non-zero coordinates.

I hope this to be true. What I have proved so far is that there is an element in $\mathbb{F}^m$ with all non-zero coordinates which is in the image of $A$ (as a linear map). I am providing some sketch of the proof what i did. Assume $W_i= \{\textbf{y}\in \mathbb{F}^n : (A\textbf{y})_{(1,i)}=0  \}.$ Then $W_i \neq \phi$   since $\textbf{0} \in W_i$ and also $W_i$ is a proper subspace of  $\mathbb{F}^n.$ For properness use the fact that $i^{th}$ row of $A$ is non-zero. If $A=(a_{ij})$ and $a_{ir_{i}} \neq 0$ then $Ae_{r_{i}}$ has $i^{th}$ coordinate non-zero where $e_i's$ form the standard basis of $\mathbb{F}^n$. Now we know that $\mathbb{F}^n \neq \bigcup \limits_{i=1}^{m}W_i.$ Then image of any element in the complement of $\bigcup \limits_{i=1}^{m}W_i$ in $\mathbb{F}^n$ has all non-zero coordinates. But we have to show that there is an element with all non-zero coordinates in the complement of   $\bigcup \limits_{i=1}^{m}W_i$. 
I need some help to prove this. Many thanks.
 A: Yes, this is true. First, we prove a special case:
Lemma. Let $ K $ be a valued field, that is, a field equipped with a nontrivial valuation. Then, the statement of the question is true for $ K $.
Proof. Equip $ K^n $ with the supremum norm coming from the ground field $ K $. Let $ V_i $ denote the subspace of $ K^n $ such that for all $ x \in V_i $, we have that the $ i $th component of $ Ax $ is zero. This subspace is a proper subspace by the rank-nullity theorem. Furthermore, let $ W_j $ denote the subspace of $ K^n $ of all vectors with $ j $th coordinate zero. Each of the $ V_i $ and the $ W_j $ are nowhere dense (this is where we need the valuation to be nontrivial) in the topological vector space $ K^n $, so their finite union is nowhere dense as well. Since a nowhere dense set cannot be all of $ K^n $, it follows that there is $ x \in K^n $ lying outside all of these sets. This $ x $ is the desired $ x $, and the proof is complete.
Now, for the trick: let $ F $ be any infinite field, then $ F(X) $ becomes a valued field under the $ X $-adic valuation, and thus the theorem is true for $ F(X) $. Let $ A $ be a matrix with nonzero rows in $ M_{m \times n}(F) $. Then, by the lemma, there is $ x \in F(X)^n $ such that no component of $ Ax $ is zero. The components of $ Ax $ are nonzero rational functions with coefficients in $ F $; and all such rational functions only have finitely many zeroes in $ F $, the same is true for the components of $ x $. Since $ F $ is infinite, there is $ y \in F $ such that each of these rational functions are nonzero at $ y $. Evaluating each coordinate of $ x $ at $ y $ then gives the desired result.
