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Consider the following matrix:

$$\left[\begin{array}{cccc} \cos(x_1-y_1) & \cos(x_1-y_2) & \ldots & \cos(x_1-y_n) \\ \cos(x_2-y_1) & \cos(x_2-y_2) & \ldots & \cos(x_2-y_n) \\ \cos(x_3-y_1) & \cos(x_3-y_2) & \ldots & \cos(x_3-y_n) \\ \vdots & \vdots & \ddots & \vdots\\ \cos(x_n-y_1) & \cos(x_n-y_2) & \ldots & \cos(x_n-y_n) \\ \end{array} \right]$$ with $x_k, y_k$ being real numbers. I cannot really apply any kind of functional calculus, I believe, to get a closed form formula for the determinant of this matrix. Any hints or ideas how to proceed?

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Let us denote by $M$ the given matrix. The answer is:

If $n \geq 3$, $\det(M)=0$ .

Proof $\sharp 1$ (using complex numbers):

Using Euler formulas, one can write

$$M=\dfrac{1}{2}(M_1+\overline{M_1})$$

with rank 1 matrix $M_1$ defined by:

$$M_1=\begin{bmatrix}e^{ix_1}\\e^{ix_2}\\\vdots\\e^{ix_n}\end{bmatrix}\begin{bmatrix}e^{-iy_1}&e^{-iy_2}&\cdots&e^{-iy_n}\end{bmatrix}$$

Thus rank$(M)\leq 2$, as sum of two rank 1 matrices (see this.)

Thus, for $n \geq 3$, by rank-nullity theorem, $dim(Ker(M))\geq 1$, and the answer follows.


Proof $\sharp 2$ (very similar proof, but using a geometrical interpretation, without complex numbers):

Let us define, in $\mathbb{R}^2$ unit vectors $U_i$ (resp. $V_{j}$) with polar angle $x_i, \ i=1,2,\cdots n $ (resp $y_j , \ j=1,2,\cdots n $).

Then $M_{i,j}=U_i \cdot V_j$ (dot-product).

Let $A$ (resp. $B$) be the $2 \times n$ matrix whose columns are the cartesian coordinates of vectors $U_i$ (resp. $V_j$).

Then $M=A^TB$, whose rank is clearly at most $2$.

Remarks:

1) If $n=1$ or $n=2$, one has in general $\det(M)\neq 0$.

2) In the case $n=2$, I don't think there is a special "closed form formula".

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