For every real vector $\mathbf{z} = (z_1,\ldots,z_N)^\top$, define the function $\textsf{sg}:\mathbb{R}^N \rightarrow \{0,1\}$ as $$ {\textsf{sg}}(\mathbf{z}) = \begin{cases} \ \ \ \phantom{-} 1 \qquad\text{if $\left(\forall\,z_i\neq 0\right)$ ${\textsf{sgn}}(z_i) = 1$ }\\[10pt] \ -1 \qquad\text{if $\left(\forall\,z_i\neq 0\right)$ ${\textsf{sgn}}(z_i) = -1$ }\\[10pt] \ \ \ \phantom{-} 0 \qquad\text{otherwise} \end{cases} $$ where $\textsf{sgn}(\cdot)$ is the usual signum function.
Conjecture: Let $\mathbf{A}$ be an $N\times N$ positive definite matrix and denote $\mathbf{A}_i$ as its $i$-th row. Then, the range of the mapping $T:[0,\infty)^N\rightarrow \mathbb{R}^N$ defined as $T: \mathbf{x} \mapsto \mathbf{A}\mathbf{x}$ is the Cartesian product $$ {\textsf{ran}}(T) = I_1 \times I_2 \times \cdots \times I_N $$ where, for $k=1,2,\ldots,N$, $$ I_k = \begin{cases} \ [0,\infty)\qquad\ \ \ \text{if $\textsf{sg}(\mathbf{A}_k) = 1$}\\[10pt] \ (-\infty,0] \qquad \text{if $\textsf{sg}(\mathbf{A}_k) = -1$}\\[10pt] \ (-\infty,\infty) \quad\ \text{if $\textsf{sg}(\mathbf{A}_k) = 0$.} \end{cases} $$ Questions: Is this true? If not, can we give one counterexample?
A special case for illustration:
Consider the positive definite matrix $$ \mathbf{A} = \begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}\text{.} $$ The mapping $\mathbf{y} = \mathbf{A}\mathbf{x}$ explicitly results to the system $$ \begin{align} y_1 &= 2x_1 - x_2 \\ y_2 &= -x_1 + 2x_2 - x_3 \\ y_3 &= -x_2 + 2x_3 \\ \end{align}\text{.} $$ Thus, by observation, one can see that as $x_1$,$x_2$, and $x_3$ takes on values from $[0,\infty)$, the conjecture implies that the values of $y_1,y_2,y_3$ ranges on the entire $\mathbb{R}$.
One can still show other positive definite matrices. The question is, is this true for all?
Suppose $\mathbf{x}\in[0,\infty)^N$. Then $T(\mathbf{x})=\mathbf{Ax}\in I_1\times I_2\times \cdots\times I_N$.
Proof: Denote the $j^\text{th}$ entry in the $i^\text{th}$ row of $\mathbf{A}$ by $\mathbf{A}_{ij}$. Then the $k^\text{th}$ entry of $T(\mathbf{x})=\mathbf{Ax}$ is $$T(\mathbf{x})_k=\sum_{i=1}^N \mathbf{A}_{ki}\mathbf{x}_i=\mathbf{A}_{k1}\mathbf{x}_1+\mathbf{A}_{k2}\mathbf{x}_2+\dots+\mathbf{A}_{N2}\mathbf{x}_2$$
It suffices to show that $T(\mathbf{x})_k\in I_k$ for all $k$.
Let $k\leq N$. If $\textsf{sg}(\mathbf{A}_k)=1$, then, by definition of signum $\mathbf{A}_{ki}>0$ for all $i\leq N$, and $I_k=[0,\infty)$. Since the domain of $T$ is $[0,\infty)^N$, each $\mathbf{x}_i\geq 0$. Thus the term $\mathbf{A}_{ki}\mathbf{x}_i\geq0$:
$$\mathbf{A}_{ki}>0\quad\text{and}\quad\mathbf{x}_i\geq0 \implies \mathbf{A}_{ki}\mathbf{x}_i\geq0$$
Since each term is $\geq 0$, the sum is $\geq 0$:
$$\forall i\leq N,\quad\mathbf{A}_{ki}\mathbf{x}_i\geq0 \implies T(\mathbf{x})_k=\sum_i \mathbf{A}_{ki}\mathbf{x}_i\geq0 \iff T(\mathbf{x})_k\in [0,\infty)=I_k$$
The argument is nearly identical if $\textsf{sg}(\mathbf{A}_k)<1$, so that $I_k=(-\infty,0]$:
$$\mathbf{A}_{ki}<0\quad\text{and}\quad\mathbf{x}_i\geq0 \implies \mathbf{A}_{ki}\mathbf{x}_i\leq0$$ $$\forall i\leq N,\quad\mathbf{A}_{ki}\mathbf{x}_i\leq0 \implies T(\mathbf{x})_k=\sum_i \mathbf{A}_{ki}\mathbf{x}_i\leq0 \iff T(\mathbf{x})_k\in (-\infty,0]=I_k$$
Otherwise, if $\textsf{sg}(\mathbf{A}_k)=0$, then $I_k=(-\infty,\infty)$, and $T(\mathbf{x})_k=\sum_i \mathbf{A}_{ki}\mathbf{x}_i \in (-\infty,\infty)$ simply because the real numbers are algebraically closed.