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.

  • 1
    $\begingroup$ What comes to mind is that, if we interchange $x_1$ and $x_2$ in the matrix, then we're effectively swapping the two columns. So whatever terrible polynomial the determinant is must be consistent with how the signs change under permutations. That suggests a linkage with (anti)symmetric polynomials in $2n$ variables. There's also a homogeneity requirement: if all the variables are rescaled by a factor $t$, then the determinant will be rescaled by a factor of $t^{nk}$. $\endgroup$ Oct 6, 2014 at 19:23
  • 4
    $\begingroup$ The matrix is the sum of $k+1$ matrices of rank $1$ so $0$ determinant for $k<n-1$. You are left with $k=n-1$. $\endgroup$
    – orangeskid
    Oct 6, 2014 at 19:31
  • $\begingroup$ @orangeskid How do you prove that ? $\endgroup$ Oct 6, 2014 at 19:50
  • 1
    $\begingroup$ @TheGame: $(\binom{k}{l} x_i^l y_j^{n-l})$ has rank $1$. $\endgroup$
    – orangeskid
    Oct 8, 2014 at 9:42
  • $\begingroup$ See arxiv.org/abs/math/9902004 $\endgroup$ Oct 8, 2014 at 19:19

3 Answers 3



For $k<n$,

$$\det(x_i+y_j)^{k}=\mathbb{1}_{k=n-1} (-1)^{\frac{n(n-1)}{2}} \prod_{k=1}^{n-1}k^{2k-n} \prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1}) $$


Fact 1

$$\det(x_i+y_j)^{k}=\mathbb{1}_{k=n-1} Const_n \prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1}) $$

where $Const_n$ is a real number that only depends on $n$ and does not depend on $x$s and $y$s.

Proof 1

If $\exists i_1, i_2: x_{i_1}=x_{i_2}$, then our determinant is $0$ because 2 rows will be the identical.

Likewise, if $\exists j_1, j_2: y_{j_1}=y_{j_2}$, then our determinant is $0$ because 2 columns will be identical.

Thus our determinant, which is clearly a polynomial, should be

$$\prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1})$$ up to a factor of some polynomial in $x,y$, i.e.

$$\det(x_i+y_j)^{n-1}=SomePolynomial(x_1,\dots, x_n, y_1, \dots, y_n)\prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1})$$

Note that $\prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1})$ is a homogeneous polynomial of degree $n(n-1)$, which puts a lower bound on the degree of the determinant.

On the other hand, if you try and write the determinant as a sum of products over all permutations with correct sign (definition), the result will be a homogeneous polynomial of degree $nk$.

Since the total degrees of these two polynomials must agree, we must have the inequality $n(n-1)\leq nk\implies n-1 \leq k$. But $k<n$ by assumption, so either $k=n-1$ or no such polynomial exists; in the latter case we conclude that the determinant vanishes identically.

If $k=n-1$, the leading factor must have degree zero i.e. the two polynomials agree up to a constant, and this constant only depends on $n$:

$$\det(x_i+y_j)^{n-1}=Const_n\prod_{i_1 < i_2}(x_{i_2}-x_{i_1})\prod_{j_1 < j_2}(y_{j_2}-y_{j_1})$$

Fact 2

$$Const_n = (-1)^{\frac{n(n-1)}{2}} \prod_{k=1}^{n-1}k^{2k-n}$$

Proof 2

If we compute the determinant according to the definition, then expand all $(x_i+y_j)^{n-1}$, we will have a linear combination of $$\prod_{i=1}^{n} x_i^{p_i} \prod_{j=1}^{n} y_j^{q_j}$$ where $\sum_{i=1}^n p_i + \sum_{j=1}^n q_j = (n-1)n$.

The contributions of element $(x_i+y_j)^{n-1}$ are of form $$\binom{n-1}{m} x_i^{m} y_j^{n-1-m} \ \forall m \in \{0,1,\dots,n-1\}$$

Thus, for a given $\prod_{i=1}^{n} x_i^{p_i} \prod_{j=1}^{n} y_j^{q_j}$, $$\exists i^* \in \{1,2,\dots,n\} : \#\{i | p_i=p_{i^*}\}=1 \implies \exists ! j^* \in \{1,2,\dots,n\} : (p_{i^*} + q_{j^*}) = n-1$$ In other words, if among the values of powers of $x$s there is a unique one, with index $i^*$, then there will be one and only one value of $q$s equal to $n-1-p_{i^*}$; we denote its index as $j^*$. This means that one of the contributors to this permutation is part of element $(x_{i^*} + y_{j^*})^{n-1}$, $\binom{n-1}{p_{i^*}} x_{i^*}^{p_{i^*}} y_{j^*}^{n-1-p_{i^*}}$

This implies that, for the terms $\prod_{i=1}^{n} x_i^{p_i} \prod_{j=1}^{n} y_j^{q_j}$ where all $p$s are unique, there is only one permutation that contributes to it. To identify this permutation, we would pair up $p$s and $q$s (all $p$s are unique, all $q$s are unique, they all belong to $\{0,1,\dots,n-1\}$, there are $n$ of each, so pairng up exists and is unique). The indices of the pairs will be the indices of the elements of $(x_i+y_j) ^{n-1}$ that contributes to the permutation, the powers will gives use the coefficients of the contribution of each element.

We will apply this fact to one of the contributions of the main diagonal,

$$x_1^{n-1} x_2^{n-2} y_2 x_3^{n-3} y_3^2 \dots x_{n-2}^{2} y_{n-2}^{n-3} x_{n-1} y_{n-1}^{n-2} y_{n}^{n-1}$$

If we the the definition of the determinant, the contribution this term receives (it will come with "+" sign), is

$\prod_{k=0}^{n-1} \binom{n-1}{k} = \frac{\left((n-1)!\right)^n}{\left(\prod_{k=0}^{n-1} k!\right)^2} = \frac{\prod_{k=1}^{n-1}k^n}{\prod_{k=1}^{n-1}k^{2(n-k)}}=\prod_{k=1}^{n-1}k^{2k-n}$

This contribution in terms of $Const_n$ is $$(-1)^{\frac{n(n-1)}{2}}Const_n$$ (all $y$s come with "+" sign, all $x$s come with "-") sign.

Thus $$Const_n = (-1)^{\frac{n(n-1)}{2}} \prod_{k=1}^{n-1}k^{2k-n}$$

Not sure if my proof is clear, but I fear that adding more details will make it even more obscure. Questions welcome!

  • $\begingroup$ $Const$ isn't $\pm 1$, as we can see in simple 3x3 examples. Can you find it ? Thanks for the answer, anyway. $\endgroup$ Oct 7, 2014 at 15:32
  • $\begingroup$ @TheGame: I will try, but if you could mark my other answer as a solution and upvote it, I will be more inclined to spend my time on it :) $\endgroup$
    – Yulia V
    Oct 7, 2014 at 16:21
  • $\begingroup$ I will don't worry, but I haven't fully looked at the other answer yet as I'm doing other things first. I'll look at it in about 1-2h $\endgroup$ Oct 7, 2014 at 16:22
  • $\begingroup$ While I agree with the intent of your approach, there are some issues. 1) I think it's clearer to talk in terms of the degree of the entire (homogeneous) polynomial rather than on a variable-by-variable basis. (The degree of the latter is bounded above by the former.) 2) Rather than 'assuming' that your symmetric polynomial in the first part, it seems simpler to take the degree of the antisymmetric part as the lower bound of the total degree. $\endgroup$ Oct 7, 2014 at 16:53
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    $\begingroup$ @YuliaV: I edited up to the $k=n-1$ case, which I don't have a good answer for. $\endgroup$ Oct 7, 2014 at 18:19

Here is an answer for the $(k,n)=(2,3)$ case which, as I will discuss at the end, can be readily generalized and made more elegant. Observe that the product of Vandermonde matrices

$$\begin{pmatrix}1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{pmatrix} \begin{pmatrix}y_1^2 & y_2^2 & y_3^2 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{pmatrix} =\begin{pmatrix} y_1^2 +x_1 y_1+x_1^2 & y_2^2 +x_1 y_2+x_1^2 & y_3^2 +x_1 y_3+x_1^2 \\ y_1^2 +x_2 y_1+x_2^2 & y_2^2 +x_2 y_2+x_2^2 & y_3^2 +x_2 y_3+x_2^2\\ y_1^2 +x_3 y_1+x_3^2 & y_2^2 +x_3 y_2+x_3^2 & y_3^2 +x_3 y_3+x_3^2 \end{pmatrix}$$

is very nearly off the desired form, with the only incongruity being that the $x_i y_j$ terms should have a coefficient of $2$. This is remedied by inserting a diagonal matrix $D=\text{diag}(1,2,1)$ between the matrices above, thereby multiplying the row of the second Vandermonde matrix by $2$. The elements of the resulting matrix are then indeed of the form $a_{ij}=(x_i+y_j)^2$.

So our matrix is the product of two Vandermonde matrices and a diagonal matrix. Recalling the form of the determinant of a Vandermonde matrix, taking the determinant thus yields the polynomial

$$-2(x_2-x_1)(x_3-x_1)(x_3-x_2)\cdot (y_2-y_1)(y_3-y_1)(y_3-y_2).$$

with a minus sign owing to the orientation of the second Vandermonde matrix.

This is a fairly nice answer, and one imagines the following generalization: For the case of $k=n-1$, the matrix is obtained by multiplying two Vandermonde matrices with a diagonal matrix of binomial coefficients $D_{ij}=\delta_{ij}\binom{n-1}{i}$. The resulting determinantal polynomial would thus be

$$-\prod_{k=1}^{n-1}\binom{n-1}{k}\prod_{1\leq j\leq k\leq n}(x_k-x_j)(y_k-y_j).$$

I imagine an elegant proof of this is possible.

  • $\begingroup$ I believe your coefficient is wrong, could you check my version? $\endgroup$
    – Yulia V
    Oct 8, 2014 at 9:45
  • $\begingroup$ My conjectured constant is definitely not quite right, though I'm not sure I agree with yours either. (Your sign is right on, though.) I'll be putting up a more general proof later with a valid final result, so I'd prefer to hold off on discussion until we can compare these. @YuliaV $\endgroup$ Oct 8, 2014 at 15:58

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

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

  • $\begingroup$ @YuliaV: Here's the answer I was alluding to in my earlier comment. The only substantial difference between our answers, I think, is the form of integer prefactor out front. I'll see if I can verify whether or not they in fact agree. $\endgroup$ Oct 8, 2014 at 19:00
  • 1
    $\begingroup$ @YuliaV: Managed to confirm that the integer prefactor agrees as well. It's equivalent to a binomial coefficients identity which I wasn't familiar with, which I posted it as a new question along with an inductive proof. So I think our answers are entirely equivalent. $\endgroup$ Oct 8, 2014 at 20:40
  • $\begingroup$ very nice proof! $\endgroup$
    – Yulia V
    Oct 8, 2014 at 21:09

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