How to evaluate $\det (1+x_iy_j)_{n\times n}$? How to evaluate 
$$\det\begin{bmatrix}1+x_1y_1&\cdots & 1+x_1y_n\\ \vdots & \ddots & \vdots \\ 1+x_ny_1 & \cdots & 1+x_ny_n\end{bmatrix}$$?
I tried factorisation but I have no idea. I could not factor anything out... I need a hint to kick off. 
Edit:
A conjecture due to multilinearity: if we let $v_j=(x_1y_j,\cdots,x_ny_j)^T$, then my guess would be
$$\det\begin{bmatrix}1+x_1y_1&\cdots & 1+x_1y_n\\ \vdots & \ddots & \vdots \\ 1+x_ny_1 & \cdots & 1+x_ny_n\end{bmatrix}=\det(1,v_2,\cdots,v_n)+\det(v_1,1,\cdots,v_n)+\cdots+\det(v_1,v_2,\cdots,1)=\color{red}{0}$$
where $1$ denote $(1,\cdots,1)^T\in \Bbb K^n$,$\Bbb K$ the underlying number field. 
 A: Hint:
$\left(\begin{matrix} 1 + x_1 y_1 & 1 + x_1 y_2 & \cdots & 1 + x_1 y_n \\ 1 + x_2 y_1 & 1 + x_2 y_2 & \cdots & 1 + x_2 y_n \\ \vdots & \vdots & \ddots & \vdots \\ 1 + x_n y_1 & 1 + x_n y_2 & \cdots & 1 + x_n y_n \end{matrix}\right)$
$= \left(\begin{matrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{matrix}\right)^T \left(\begin{matrix} 1 & 1 & \cdots & 1 \\ y_1 & y_2 & \cdots & y_n \\ 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{matrix}\right)$
(where both matrices on the right hand side are $n\times n$-matrices).
A: if $n = 1,$  then the determinant is $1 + x_1y_1$ and for $n = 2,$ it is $(x_1-x_2)(y_1 - y_2).$  
let $1$ be the column vector with all components equal to $1.$
i think the determinant of $A=11^\top + xy^\top$ for $n \ge 3$ is zero. the reason is $0$ is an eigenvalue of $A.$ in dimension $n \ge 3,$ we can choose a nonzero vector $u$  such that $u$ is orthogonal to both $1$ and $y.$  we have $Au = (11^\top + xy^\top)u = (1^\top u)1+(y^\top u) x = 0.$ 
A: Another way of seeing that the determinant is $0$ for $n>2$ is as follows:
The determinant is a polynomial in $x_1$. If $n\geq 3$ then $x_2$ and $x_3$ are roots of this polynomial. (Since if $x_1=x_i$ then we have two rows which are equal)
Thus $(x_1-x_2)(x_1-x_3)$ is a factor of the determinant.
But we can also see that the determinant should be linear in $x_1$ since $x_1$ only occurs in the first row.
This implies that the determinant must be $0$. (Otherwise we would have a linear polynomial with a quadratic factor)
A: Hints:


*

*$rank(A+B)\leq rank(A)+rank(B)$

*$rank(1_{n\times m})=1$

*$rank(xy^t)=1$

