Let $A$ be a symmetric square matrix with entries in $\mathbb{Z}/p\mathbb{Z}$ for a prime $p$ such that all of its diagonal entries are nonzero. Does there exists always a vector $x$ with all coordinates nonzero in the image of $A$? (this is true for $p=2$, but I don't know the answer for other values of $p$)



A linear algebra problem in positive characteristic
This is false. Let $c$ be a quadratic nonresidue modulo $p$. Our matrix will be $(p^2-1) \times (p^2-1)$, with rows and colums indexed by pairs $(x,y) \in \mathbb{F}\_p^2 \setminus \{ (0,0) \}$. 


Our matrix is defined by
$$A\_{(x\_1,y\_1) \ (x\_2, y\_2)} = x\_1 x\_2 - c y\_1 y\_2.$$
This is obviously symmetric. Since $c$ is a nonresidue, we have $x^2-cy^2 \neq 0$ for $(x,y) \in \mathbb{F}\_p^2 \setminus \{ (0,0) \}$, so the diagonal entries are nonzero. 


Each column of this matrix is a linear function of $(x,y)$. So every vector in the image of this matrix is a linear function $\mathbb{F}\_p^2 \setminus \{ (0,0) \} \longrightarrow \mathbb{F}\_p$ and, hence, takes the value $0$ somewhere.


We could make a smaller $(p+1) \times (p+1)$ example by just taking one point $(x,y)$ on each line through $0$ in $\mathbb{F}\_p^2$.




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Moreover, I claim that $(p+1) \times (p+1)$ is optimal. In other words, if $A$ is an $n \times n$ matrix with $n \leq p$ and nonzero entries on the diagonal, then some vector in the image of $A$ has all coordinates nonzero. Interestingly, I don't need the symmetry hypothesis.


Let $W$ be the image of $A$. Note that $W$ is not contained in any of the coordinate hyperplanes. 


Let $\vec{u}= (u\_1,\ u\_2, \ \ldots,\ u\_n)$, among all elements of $W$, have the fewest $0$ entries. Suppose for the sake of contradiction that some $u\_i$ is $0$. Then there is some other $\vec{v}$ in $W$ with $v\_i \neq 0$. Consider the points $(u\_j : v\_j)$ in $\mathbb{P}^1(\mathbb{F}\_p)$ as $j$ ranges over all indices where $(u\_j, v\_j) \neq (0,0)$. There are fewer then $p+1$ such $j$, so some point of $\mathbb{P}^1(\mathbb{F}\_p)$ is not hit, call it $(a:b)$. Then $-b \vec{u} + a \vec{v}$ is in $W$ and has fewer nonzero entries than $\vec{u}$, a contradiction.


