How to prove the equation $x^TAy=v$ has a solution? Problem:
Given a matrix $A$ with shape $n\times n$ and a scalar value $v$, consider the equation system
$$x^TAy = v,\\
\mathbf{1}_n^Tx=1, x\geq \mathbf{0}\\
\mathbf{1}_n^Ty=1, y\geq \mathbf{0}$$
where $x, y$ are in a $n-1$ probability simplex.
Consider a simplest case
Given a matrix $A=\begin{bmatrix}
     a_{1} & a_{2} \\
     a_{3} & a_{4} \\
     \end{bmatrix}$ and $v$, the decision variable are $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}$, $y=\begin{bmatrix}y_1\\y_2\end{bmatrix}$.
the problem can be written as
$$a_1x_1y_1+a_2x_1y_2+a_3x_2y_1+a_4x_2y_2=v\\
x_1+x_2=1, x\geq \mathbf{0}\\
y_1+y_2=1, y\geq \mathbf{0}$$
Some of my tries
I actually put it to Gurobi directly without any reformulation, and the solver always gives a correct solution, so I think the existence of a solution can be proved.
What is such a problem called? and what method can solve it? Any idea to prove that there always exists at least one pair of $(x, y)$ satisfy the above equation system?
 A: The inequalities
$$
\begin{array}{rl}
\mathbf{1}_n^Tx =1, &x\geq \mathbf{0}\\
\mathbf{1}_n^Ty =1, &y\geq \mathbf{0}
\end{array}
$$
implies $0\leq x_{i}\leq 1$, $0\leq y_{j}\leq 1$,
$$
x_{1}+\ldots+x_{n}=1 \quad \mbox{ and } \quad y_{1}+\ldots+y_{n}=1.
$$
These inequalities make it easy to proof that
$$
\min_{ij}A_{ij}\leq x^{T}Ay=\sum_{i=1}\sum_{j=1}x_{i}\cdot A_{ij}\cdot y_{j}\leq \max_{ij}A_{ij}
$$
Then the equation
$$
x^{T}Ay=\sum_{i=1}\sum_{j=1}x_{i}\cdot A_{ij}\cdot y_{j}=v
$$
has solution if, only if,  $\min_{ij}A_{ij}\leq v \leq \max_{ij}A_{ij}$. In fact, supose $A_{\alpha,\beta}=\min_{ij}A_{ij}$ and $A_{\mu,\nu}=\max_{ij}A_{ij}$ set $y_{\beta}=t$, $y_{\nu}=1-t$, $x_{\alpha}=t$ and $x_{\mu}=1-t$ for $0\leq t\leq 1$. Then
$$
x^{T}Ay
=
t\cdot A_{\alpha \beta}\cdot t+(1-t)\cdot A_{\mu \nu}\cdot (1-t)
=
t^{2}\cdot A_{\alpha \beta}+(1-t)^{2}\cdot A_{\mu \nu}.
$$
Now just note that the quadratic equation
$$
t^{2}\cdot A_{\alpha \beta}+(1-t)^{2}\cdot A_{\mu \nu}=v
$$
has solution in the $[0,1]$ interval when $A_{\alpha \beta}\leq v\leq  A_{\mu \nu}$.
