TITLE: Determinant of $a_{i,j}=(x_i+y_j)^k$
QUESTION [11 upvotes]: How can I find the determinant of the matrix $A\in\mathcal{M}_n(\mathbb{R})$ with coefficients $a_{i,j}=(x_i+y_j)^k,k<n$ ?
All the $x_u,y_u$ are real numbers.
Derivating won't help, and I didn't find any good way to simplify the problem.

REPLY [2 votes]: Here is a refinement/generalization of the approach in my first answer. First, note that by way of the binomial theorem we may express $(x+y)^k$ for given $x,y$ in bilinear form:
\begin{align}
(x+y)^k
=\sum_{l=0}^k\binom{k}{l}x^l y^{k-l}
&=\begin{pmatrix}1 & x & \cdots & x^{k-1} & x^k \end{pmatrix}\\
&\quad \cdot \text{diag}\left(1,k,\binom k2,\cdots,k,1 \right)\\
&\quad \cdot \begin{pmatrix}y^k & y^{k-1} & \cdots & y & 1\end{pmatrix}^T
\end{align}
We can write this compactly as $(x+y)^k = \mathbf{x}^T (DR) \mathbf{y} $ with $(k+1)\times 1$ column vectors  $(\mathbf{x})_{0\leq i \leq  k}=x^i$, $(\mathbf{y})_{0\leq i \leq  k}=y^i$, a $(k+1)\times (k+1)$ diagonal matrix $D=\text{diag}\left[\binom{k}{i}\right]_{0\leq i \leq k}$, and $R$ the $(k+1)\times (k+1)$ order-reversing permutation matrix $(R)_{0\leq i,j\leq k}=\delta_{i,k-j}$. 
Applying this formula element-wise to the problem of interest, we have  $a_{ij}=(x_i+y_j)^k=\mathbf{x}_i^T (DR)\mathbf{y}_j$ and so $A=X^T( DR )Y$ where we have introduced the $(k+1)\times n$ rectangular matrices $$X=\begin{pmatrix}\mathbf{x}_1 & \mathbf{x}_2 & \cdots  & \mathbf{x}_n\end{pmatrix},\\ Y=\begin{pmatrix}\mathbf{y}_1 & \mathbf{y}_2 & \cdots  & \mathbf{y}_n\end{pmatrix}.$$
Writing this as $X^T(DRY)$, we have reduced our problem to that of finding the determinant of a product of $(k+1)\times n$ and $n\times (k+1)$ rectangular matrices. There are three cases:

If $k+1 > n$, then the LH matrix has more rows than columns; but this is ruled out by the assumption $k<n$. (I'm inclined to post a new question for this case.)
If $k+1<n$, then the LH matrix has more columns than rows. Consequently the resulting matrix cannot have full rank (specifically, it has rank at most $n$) and the determinant vanishes.
If $k+1=n$, then all the matrices are square and the determinant of the product is the product of the determinants. For $X^T$ and $Y$, we observe that these are (up to a matrix transpose) Vandermonde matrices over the variables $\{x_{1\leq i \leq n}\}$ and $\{y_{1\leq i \leq n}\}$; the corresponding Vandermonde determinants are $\prod_{1\leq i< j\leq n}(x_j-x_i)$ and $\prod_{1\leq i< j\leq n}(y_j-y_i)$. The diagonal matrix $D$ has determinant $\prod_{i=0}^{n-1}\binom{n-1}{i}$. Since the order-reversing matrix $R$ has may be converted to the identity matrix by $\lfloor n/2 \rfloor$ row transpositions, we have an overall sign of $(-1)^{\lfloor n/2\rfloor}$. (For comparison with Yulia's answer, this may also be written as $(-1)^{n(n-1)/2}$ since both exponents have the same parity over integers.) Multiplying these factors, we have the final result for the $k=n-1$ case as

$$\displaystyle \boxed{(-1)^{\lfloor n/2\rfloor}\prod\limits_{i=0}^{n-1}\binom{n-1}{i}\prod\limits_{1\leq i< j\leq n}(x_j-x_i)(y_j-y_i)}$$