Vandermonde Determinants

A similar formula holds when the monomials $x^i$ are replaced by an arbitrary polynomial basis $P_i(x)$ of the $n$ dimensional vector space of polynomials of degree less than $n$.

This formula follows directly from equation vandermonde by writing the LHS of basis as the product of the two matrices $c_{ij}$ and $x_i^j$ and taking the determinant.

If we knock out one row and one column from a Vandermonde matrix we can still compute it's determinant with a formula very similar to those above, for example

\begin{equation*} \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^4 \\ 1 & x_2 & x_2^2 & x_2^4 \\ 1 & x_3 & x_3^2 & x_3^4 \\ 1 & x_4 & x_4^2 & x_4^4 \\ \end{vmatrix} \space = \space \left(x_1+x_2+x_3+x_4\right) \cdot \prod\limits_{i\lt j} (x_j-x_i) \end{equation*}

This identity follows by expanding each side of the usual Vandermonde determinant formula as a polynomial in $t$ and equating coefficients.

The LHS of minormiracle is the coefficient of $-t^3$ on the LHS of proof and the RHS of minormiracle is the coefficient of $-t^3$ on the RHS of proof.

\begin{equation*} \begin{vmatrix} 1 & x_1 & x_1^3 & x_1^4 \\ 1 & x_2 & x_2^3 & x_2^4 \\ 1 & x_3 & x_3^3 & x_3^4 \\ 1 & x_4 & x_4^3 & x_4^4 \\ \end{vmatrix} \space = \space \left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\right) \cdot \prod\limits_{i\lt j} (x_j-x_i) \end{equation*}

\begin{equation*} \begin{vmatrix} 1 & x_1^2 & x_1^3 & x_1^4 \\ 1 & x_2^2 & x_2^3 & x_2^4 \\ 1 & x_3^2 & x_3^3 & x_3^4 \\ 1 & x_4^2 & x_4^3 & x_4^4 \\ \end{vmatrix} \space = \space \left(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4\right) \cdot \prod\limits_{i\lt j} (x_j-x_i) \end{equation*}

Form an $n \times n$ matrix whose rows are the elementary symmetric polynomials on $n-1$ variables, the $i$-th row having $x_i$ omitted. Call this the dual Vandermonde matrix, then it has the same determinant as the usual Vandermonde matrix:

\begin{equation*} \begin{vmatrix} x_2x_3x_4 & x_2x_3 + x_2x_4 + x_3x_4 & x_2 + x_3 + x_4 & 1 \\ x_1x_3x_4 & x_1x_3 + x_1x_4 + x_3x_4 & x_1 + x_3 + x_4 & 1 \\ x_1x_2x_4 & x_1x_2 + x_1x_4 + x_2x_4 & x_1 + x_2 + x_4 & 1 \\ x_1x_2x_3 & x_1x_2 + x_1x_3 + x_2x_3 & x_1 + x_2 + x_3 & 1 \\ \end{vmatrix} \enspace = \enspace \prod\limits_{i \lt j}(x_j - x_i) \end{equation*}

Follows from the fact that the determinant is an homogenous polynomial of degree $\tfrac 1 2 n(n-1)$ which vanishes when $x_i = x_j$ for $i \ne j$, because then the $i$-th and $j$-th rows are identical. Also dividing the first $n-1$ columns on the LHS by $x_n$ and dividing the $n-1$ factors like $(x_n-x_i)$ on the RHS by $x_n$ and letting $x_n \rightarrow \infty$ converts the $n$-th formula into the $(n-1)$-th formula. And because the formula for $n=1$ evaluates to $1$, the unknown constant on the RHS is $1$.

The $n \times n$ matrix $m_{ij} = \sum\limits_{k,l} c_{kl} \thinspace x_i^k \thinspace y_j^l$ of polynomials factors into the product of three matrices and its determinant is given by

\begin{equation*} \det\left(m_{ij}\right) \space = \space \det\left(c_{ij}\right) \cdot \prod\limits_{i\lt j} (x_j-x_i) \cdot \prod\limits_{i\lt j} (y_j-y_i) \end{equation*}

This result follows from taking the determinant of the matrix factorisation

$\displaystyle (m_{ij}) \space = \space \begin{pmatrix} 1 & x_1 & \ldots & x_1^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & \ldots & x_n^{n-1} \\ \end{pmatrix} \allowbreak \begin{pmatrix} c_{11} & c_{12} & \ldots & c_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \ldots & c_{nn} \\ \end{pmatrix} \allowbreak \begin{pmatrix} 1 & y_1 & \ldots & y_1^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & y_n & \ldots & y_n^{n-1} \\ \end{pmatrix}^T $

This is easier to see with tensor index notation

\begin{equation*} M_i^j \space = \space X_i^k \thinspace C_k^l \thinspace Y_l^j \end{equation*}

where the components of the tensors are given by $M_i^j = m_{ij}, \enspace X_i^k = x_i^k, \enspace C_k^l = c_{kl}, \enspace Y_l^j = y_j^l$

When $c_{ij}$ is the matrix with binomial coefficients down the anti-diagonal and zeroes elsewhere we get

\begin{equation*} \begin{vmatrix} (x_1+y_1)^{n-1} & (x_1+y_2)^{n-1} & \cdots & (x_1+y_n)^{n-1} \\ (x_2+y_1)^{n-1} & (x_2+y_2)^{n-1} & \cdots & (x_2+y_n)^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ (x_n+y_1)^{n-1} & (x_n+y_2)^{n-1} & \cdots & (x_n+y_n)^{n-1} \\ \end{vmatrix} \space = \space (-1)^{\sfrac 1 2 n(n-1)} \cdot \prod\limits_{i=0}^{n-1}{n-1 \choose i} \cdot \prod\limits_{i\lt j} (x_j-x_i) \cdot \prod\limits_{i\lt j} (y_j-y_i) \end{equation*}

\begin{equation*} \begin{vmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & (x_1+y_1)^n & (x_1+y_2)^n & \cdots & (x_1+y_n)^n \\ 1 & (x_2+y_1)^n & (x_2+y_2)^n & \cdots & (x_2+y_n)^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & (x_n+y_1)^n & (x_n+y_2)^n & \cdots & (x_n+y_n)^n \\ \end{vmatrix} \space = \space (-1)^{\sfrac 1 2 n(n+1)} \cdot \prod\limits_{i=0}^{n}{n \choose i} \cdot \prod\limits_{i\lt j} (x_j-x_i) \cdot \prod\limits_{i\lt j} (y_j-y_i) \end{equation*}

In the $n + 1$ case of dblvan1, put $x_{n+1} = y_{n+1} = t$, divide each side by $t^{2n}$, let $t \rightarrow \infty$ and finally move the last column and row to the first column and row position.

Multiplying a Vandermonde matrix in $x_i$ by the transpose of a Dual Vandermonde matrix in $y_i$ gives

\begin{equation*} \begin{vmatrix} (x_1+y_1)^{-1} & (x_1+y_2)^{-1} & \cdots & (x_1+y_n)^{-1} \\ (x_2+y_1)^{-1} & (x_2+y_2)^{-1} & \cdots & (x_2+y_n)^{-1} \\ \vdots & \vdots & \ddots & \vdots \\ (x_n+y_1)^{-1} & (x_n+y_2)^{-1} & \cdots & (x_n+y_n)^{-1} \\ \end{vmatrix} \space = \space \frac {\prod\limits_{i\lt j} (x_j-x_i) \cdot \prod\limits_{i\lt j} (y_j-y_i)} {\prod\limits_{i,j} (x_i+y_j)} \end{equation*}

The inner product of the $i$-th row of the vandermonde matrix in $x_i$, denoted by $X$, with the $j$-th row of the dual vandermonde matrix in $y_i$, denoted by $Y$, is given by

$\displaystyle \left(XY^T\right)_{ij} \space = \space \sum_{k=0}^n x_i^k \thinspace e_{n-j}(S_j) \space = \space \prod\limits_{y \in S_j} (x_i + y) \space = \space (x_i+y_j)^{-1} \prod_{k=1}^n (x_i+y_k) $

where $S_j = \left\{y_k: 1 \le k \le n, k \ne j\right\}$ and $e_m(S_j)$ is the elementary symmetric polynomial of degree $m$ on the $n-1$ variables $S_j$. Therefore

$\displaystyle X Y^T \space = \space \begin{pmatrix} (x_1+y_1)^{-1} \prod\limits_{k=1}^n (x_1+y_k) & \cdots & (x_1+y_n)^{-1} \prod\limits_{k=1}^n (x_1+y_k) \\ \vdots & \ddots & \vdots \\ (x_n+y_1)^{-1} \prod\limits_{k=1}^n (x_n+y_k) & \cdots & (x_n+y_n)^{-1} \prod\limits_{k=1}^n (x_n+y_k) \\ \end{pmatrix} \space = \space \begin{pmatrix} (x_1+y_1)^{-1} & \cdots & (x_1+y_n)^{-1} \\ \vdots & \ddots & \vdots \\ (x_n+y_1)^{-1} & \cdots & (x_n+y_n)^{-1} \\ \end{pmatrix} \cdot \prod\limits_{l=1}^n \prod\limits_{k=1}^n (x_l+y_k) $

and dblvan2 follows on taking determinants of both sides.

\begin{equation*} \begin{vmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & (x_1+y_1)^{-1} & (x_1+y_2)^{-1} & \cdots & (x_1+y_n)^{-1} \\ 1 & (x_2+y_1)^{-1} & (x_2+y_2)^{-1} & \cdots & (x_2+y_n)^{-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & (x_n+y_1)^{-1} & (x_n+y_2)^{-1} & \cdots & (x_n+y_n)^{-1} \\ \end{vmatrix} \space = \space - \frac {\sum\limits_i (x_i + y_i) \cdot \prod\limits_{i\lt j} (x_j-x_i) \cdot \prod\limits_{i\lt j} (y_j-y_i)} {\prod\limits_{i,j} (x_i + y_j)} \end{equation*}

This formula can be obtained by computing the power series expansion in $t$ of dblvan2 at $n+1$ with $x_{n+1} = y_{n+1} = 1/t$. Using

\begin{equation*} \frac 1 {x + 1/t} \enspace = \enspace t \space - \space xt^2 + \space \bigO(t^3) \end{equation*}

the left hand side of dblvan2 at $n+1$ is

\begin{equation*} \begin{vmatrix} (x_1+y_1)^{-1} & \cdots & (x_1+y_n)^{-1} & t + \bigO(t^2) \\ (x_2+y_1)^{-1} & \cdots & (x_2+y_n)^{-1} & t + \bigO(t^2) \\ \vdots & \ddots & \vdots & \vdots \\ t + \bigO(t^2) & \cdots & t + \bigO(t^2) & \tfrac 1 2 t \\ \end{vmatrix} \enspace = \enspace \begin{vmatrix} (x_1+y_1)^{-1} & \cdots & (x_1+y_n)^{-1} & 0 \\ (x_2+y_1)^{-1} & \cdots & (x_2+y_n)^{-1} & 0 \\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 0 & \tfrac 1 2 \\ \end{vmatrix} \thinspace t \enspace + \enspace \begin{vmatrix} (x_1+y_1)^{-1} & \cdots & (x_1+y_n)^{-1} & 1 \\ (x_2+y_1)^{-1} & \cdots & (x_2+y_n)^{-1} & 1 \\ \vdots & \ddots & \vdots & \vdots \\ 1 & \cdots & 1 & 0 \\ \end{vmatrix} \thinspace t^2 \enspace + \enspace \bigO(t^3) \end{equation*}

the right hand side of dblvan2 at $n+1$ is

\begin{equation*} \frac t 2 \prod_{i=1}^n \frac {(1 - tx_i)(1 - ty_i)} {(1 + tx_i)(1 + ty_i)} \cdot \Delta \enspace = \enspace \frac t 2 \left[1 \enspace - \enspace 2t\sum_{i=1}^n (x_i + y_i) \enspace + \enspace \bigO(t^2) \right] \cdot \Delta \end{equation*}

where $\Delta$ is the right hand side of dblvan2 at $n$. The result follows by equating the coefficients of $t^2$ terms.

There is another simple depleted determinant formula that I have come across while investigating elliptic curves.

Let $M$ be a square matrix of order $n$ and let $\minors{M}$ denote the matrix of minors of $M$. Let overbar denote a sub-matrix operation which selects $m$ rows and columns and underbar denote the complementary sub-matrix operation which selects the remaining $n - m$ rows and columns. Then

\begin{equation*} \det(\overline{\minors{M}}) \enspace = \enspace \det(M)^{m-1} \det(\underline{M}) \end{equation*}

I have verified subdet using CAS for $n=3,4,5$ presumably a proof goes something like this 🤔

We know that if $\det(M) = 0$ then $\rank(\minors{M}) \le 1$.
Assume $\rank(\minors{M}) = 1$ then all rows of $\minors{M}$ are a multiple of each other and no row can be all zeroes.
The determinant of any square $2\times 2$ submatrix must be zero otherwise we would have two linearly independent rows.
Therefore the algebraic expression for the determinant of each $2\times 2$ submatrix is divisible by $\det(M)$.
We can write the determinant of a $3\times 3$ submatrix as a linear sum of $2\times 2$ determinants which we can therefore divide by $\det(M)$.
If the resulting expression is non-zero we can again construct two linearly independent rows. Therefore the algebraic expression for the determinant of any $3\times 3$ submatrix is divisible by $\det(M)^2$.
And by induction the determinant of any square $m\times m$ submatrix is divisible by $\det(M)^{m-1}$.
If $\det(\overline{\minors{M}}) = 0$ and $\det(M) \ne 0$ the determinant of the excluded rows and columns must be zero.
Thus the RHS is determined up to an unknown multiplicative constant. By setting $M$ to a diagonal matrix that constant can be determined to be 1.

and the sub-matrix operation which selects rows 1 and 2, and columns 1 and 2 of $\minors{M}$ we have

\begin{equation*} \begin{vmatrix} b_2 & c_2 & d_2 \\ b_3 & c_3 & d_3 \\ b_4 & c_4 & d_4 \\ \end{vmatrix} \space \begin{vmatrix} a_1 & c_1 & d_1 \\ a_3 & c_3 & d_3 \\ a_4 & c_4 & d_4 \\ \end{vmatrix} \enspace - \enspace \begin{vmatrix} b_1 & c_1 & d_1 \\ b_3 & c_3 & d_3 \\ b_4 & c_4 & d_4 \\ \end{vmatrix} \space \begin{vmatrix} a_2 & c_2 & d_2 \\ a_3 & c_3 & d_3 \\ a_4 & c_4 & d_4 \\ \end{vmatrix} \enspace = \enspace \begin{vmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \\ \end{vmatrix} \space \begin{vmatrix} c_3 & d_3 \\ c_4 & d_4 \\ \end{vmatrix} \end{equation*}

This can also be written as a formula for the determinant of a sub-matrix of the inverse of $M$

Vandermonde determinants also give a formula for the discriminant of a polynomial in terms of the sums of powers of roots.

Let $P(x) = (x-e_1)(x-e_2)(x-e_3)(x-e_4)$ and $s_k = \sum e_i^k$ then using the product formula for determinants we get

\begin{equation*} \discrim(P) \space = \space \prod_{i \lt j} (e_i - e_j)^2 \space = \space \begin{vmatrix} 1 & e_1 & e_1^2 & e_1^3 \\ 1 & e_2 & e_2^2 & e_2^3 \\ 1 & e_3 & e_3^2 & e_3^3 \\ 1 & e_4 & e_4^2 & e_4^3 \\ \end{vmatrix}^2 \space = \space \begin{vmatrix} s_0 & s_1 & s_2 & s_3 \\ s_1 & s_2 & s_3 & s_4 \\ s_2 & s_3 & s_4 & s_5 \\ s_3 & s_4 & s_5 & s_6 \\ \end{vmatrix} \end{equation*}

The Laplace expansion of a determinant by cofactors of the first column can be written like this (writing $.$ instead of zero for clarity)

\begin{aligned} \begin{vmatrix} a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4 \\ d_1 & d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace &= \enspace \begin{vmatrix} a_1 & . & . & . \\ . & b_2 & b_3 & b_4 \\ . & c_2 & c_3 & c_4 \\ . & d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} . & a_2 & a_3 & a_4 \\ b_1 & . & . & . \\ . & c_2 & c_3 & c_4 \\ . & d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} . & a_2 & a_3 & a_4 \\ . & b_2 & b_3 & b_4 \\ c_1 & . & . & . \\ . & d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} . & a_2 & a_3 & a_4 \\ . & b_2 & b_3 & b_4 \\ . & c_2 & c_3 & c_4 \\ d_1 & . & . & . \\ \end{vmatrix} \\\\ &= \enspace a_1 \begin{vmatrix} b_2 & b_3 & b_4 \\ c_2 & c_3 & c_4 \\ d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace - \enspace b_1 \begin{vmatrix} b_2 & b_3 & b_4 \\ c_2 & c_3 & c_4 \\ d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace c_1 \begin{vmatrix} a_2 & a_3 & a_4 \\ b_2 & b_3 & b_4 \\ d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace - \enspace d_1 \begin{vmatrix} a_2 & a_3 & a_4 \\ b_2 & b_3 & b_4 \\ c_2 & c_3 & c_4 \\ \end{vmatrix} \\ \end{aligned}

In a similar way the determinant can be expanded by sub-determinants of the first two columns like this

\begin{aligned} \begin{vmatrix} a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4 \\ d_1 & d_2 & d_3 & d_4 \\ \end{vmatrix} \enspace &= \enspace \begin{vmatrix} a_1 & a_2 & . & . \\ b_1 & b_2 & . & . \\ . & . & c_3 & c_4 \\ . & . & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} a_1 & a_2 & . & . \\ . & . & b_3 & b_4 \\ c_1 & c_2 & . & . \\ . & . & d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} a_1 & a_2 & . & . \\ . & . & b_3 & b_4 \\ . & . & c_3 & c_4 \\ d_1 & d_2 & . & . \\ \end{vmatrix} \enspace + \enspace \textsf{3 more terms} \\\\ &= \enspace \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{vmatrix} \begin{vmatrix} c_3 & c_4 \\ d_3 & d_4 \\ \end{vmatrix} \enspace - \enspace \begin{vmatrix} a_1 & a_2 \\ c_1 & c_2 \\ \end{vmatrix} \begin{vmatrix} b_3 & b_4 \\ d_3 & d_4 \\ \end{vmatrix} \enspace + \enspace \begin{vmatrix} a_1 & a_2 \\ d_1 & d_2 \\ \end{vmatrix} \begin{vmatrix} b_3 & b_4 \\ c_3 & c_4 \\ \end{vmatrix} \enspace + \enspace \textsf{3 more terms} \\ \end{aligned}

then the determinant of $M$ is given by summing over all the $n \choose k$ possible sub-matrices $L$

When $k = 1$ this is the Laplace formula for computing the determinant using cofactors of the first column.

Given two algebraic curves $F(x,y) = 0$ and $G(x,y) = 0$ their resultant with respect to $y$ is a polynomial $\mathfrak{R}(x)$ that can be written

\begin{equation*} \mathfrak{R}(x) \space = \space \resultant_y(F,G) \space = \space \operatorname{lead}_y(F)^n \operatorname{lead}_y(G)^m \prod_{i=1}^m \prod_{j=1}^n (\psi_i(x) - \phi_j(x)) \space = \space \operatorname{lead}_y(F)^n \prod_{i=1}^m G(x,\psi_i(x)) \space = \space (-1)^{mn} \operatorname{lead}_y(G)^m \prod_{i=1}^n F(x,\phi_i(x)) \end{equation*}

where $m=\deg_y(F)$, $n=\deg_y(G)$ and $\psi_i(x)$ are the $m$ roots of $F(x,\psi_i(x)) = 0$ and $\phi_i(x)$ are the $n$ roots of $G(x,\phi_i(x)) = 0$.

\begin{equation*} F\left( x_3 \begin{vmatrix} x_1 & z_1 \\ x_2 & z_2 \\ \end{vmatrix}, \space x_3 \begin{vmatrix} y_1 & z_1 \\ y_2 & z_2 \\ \end{vmatrix} \space + \space z_3 \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \\ \end{vmatrix}, \space z_3 \begin{vmatrix} x_1 & z_1 \\ x_2 & z_2 \\ \end{vmatrix} \right) \enspace = \enspace z_1^2 z_2^2 \begin{vmatrix} x_1 & z_1 \\ x_3 & z_3 \\ \end{vmatrix} \begin{vmatrix} x_2 & z_2 \\ x_3 & z_3 \\ \end{vmatrix} \begin{vmatrix} X & Z \\ x_3 & z_3 \\ \end{vmatrix} \enspace - \enspace \begin{vmatrix} x_2 & z_2 \\ x_3 & z_3 \\ \end{vmatrix}^3 F(x_1,y_1,z_1) \enspace + \enspace \begin{vmatrix} x_1 & z_1 \\ x_3 & z_3 \\ \end{vmatrix}^3 F(x_2,y_2,z_2) \end{equation*}