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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.

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  • $\begingroup$ Is there a reason you expect there is a simple form? $\endgroup$ – Thomas Andrews Sep 7 '15 at 1:39
  • $\begingroup$ @ThomasAndrews Yes it is an exercise. $\endgroup$ – Vim Sep 7 '15 at 1:41
  • $\begingroup$ so, dude, what is the rank? $\endgroup$ – Will Jagy Sep 7 '15 at 1:51
  • $\begingroup$ @WillJagy I don't know.. But if my conjecture is right, it should be singular. $\endgroup$ – Vim Sep 7 '15 at 1:54
  • $\begingroup$ Try row operations? $\endgroup$ – Empy2 Sep 7 '15 at 1:55
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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)

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  • $\begingroup$ Nicest approach so far. +1. $\endgroup$ – Vim Sep 7 '15 at 2:12
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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).

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  • $\begingroup$ brilliant decomposition, +1. $\endgroup$ – Vim Sep 7 '15 at 2:05
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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 $Au = (11^\top + xy^\top)u = (1^\top u)1+(y^\top u) x = 0.$

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  • $\begingroup$ Excellent answer. It's neater than my multi-linearity approach. $\endgroup$ – Vim Sep 7 '15 at 2:01
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Hints:

  • $rank(A+B)\leq rank(A)+rank(B)$
  • $rank(1_{n\times m})=1$
  • $rank(xy^t)=1$
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  • $\begingroup$ Could you explain more? $\endgroup$ – user444042 Nov 23 '17 at 17:24

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