# Determinant (and invertibility) of generalized Vandermonde matrix

I have stumbled upon the following generalization of Vandermonde matrix when solving some problem in linear algebra related to Jordan normal form.

Let us consider some number $\lambda$ and we assign to this number an $n\times m$ matrix $V_m(\lambda)$ such that the first column is of the form $(1,\lambda,\lambda^2,\dots,\lambda^{n-1})^T$, the second column is of the form $(0,1,2\lambda,\dots,(n-1)\lambda^{n-2})^T$ etc. I.e., the $m$-th column will be $(0,\dots,0,1,\binom{m}{m-1}\lambda,\dots,\binom{n-1}{m-1}\lambda^{n-m})$, i.e. $$V_m(\lambda)= \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ \lambda & 1 & 0 & \ldots & 0 \\ \lambda^2 & 2\lambda & 1 & \ldots & 0 \\ \lambda^3 & 3\lambda^2 & 3\lambda & \ldots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ \lambda^{n-1} & (n-1)\lambda^{n-2} & \binom{n-1}2\lambda^{n-3} & \ldots & \binom{n-1}{m-1}\lambda^{n-m} \end{pmatrix}$$ In the other words, the entry in $k$-th row and $l$-th column is $\binom{k-1}{l-1}x^{k-l}$.

Now if we have some numbers $m_1,\dots,m_k$ such that $m_1+\dots+m_k=n$, we can define an $n\times n$-matrix $$V_{m_1,\dots,m_k}(\lambda_1,\dots,\lambda_k)= \begin{pmatrix}V_{m_1}(\lambda_1) & V_{m_2}(\lambda_2) & \dots & V_{m_k}(\lambda_k) \end{pmatrix}.$$

For example, $$V_{3,2}(x,y)= \begin{pmatrix} 1 & 0 & 0 & 1 & 0 \\ x & 1 & 0 & y & 1 \\ x^2 & 2x & 1 & y^2 & 2y \\ x^3 & 3x^2 & 3x & y^3 & 3y^2 \\ x^4 & 4x^3 & 6x & y^4 & 4y^3 \end{pmatrix}$$

Such matrix is indeed called generalized Vandermonde matrix by some authors, for example here or here. (Although the term generalized Vandermonde matrix is also used in different meanings, for example here.)

The determinant of generalized Vandermonde matrix is $$\prod_{i<j} (\lambda_j-\lambda_i)^{m_im_j}.$$

We already have at this site several questions about the usual Vandermonde matrix, for example Vandermonde Determinant, Vandermonde determinant by induction, Proof determinant of transpose Vandermonde matrix is $\prod_{1\le i\lt j\le n}(\alpha_i-\alpha_j)$, Why are Vandermonde matrices invertible?

Various derivations of determinant of Vandermonde matrix and also some proofs of the fact that it is invertible (for distinct $\lambda_i$'s) are given in those questions. I am wondering about the same question for generalized Vandermonde matrix.

How can we show that generalized Vandermonde matrix is invertible when $\lambda_i\ne\lambda_j$? How can we evaluate the determinant of generalized Vandermonde matrix?

This is a proof of the invertibility of the matrix only, when the $\lambda_i$'s are distinct.

Let $a_0,a_1,\ldots,a_{n-1}$ be coefficients, such that the corresponding linear combination of the rows of $V_{m_1,\ldots,m_k}(\lambda_1,\ldots,\lambda_k)$ is zero.

Denote by $P$ the following polynomial:

$$P=a_0+a_1x+\cdots+a_{n-1}x^{n-1}$$

The assumption is equivalent to the following equalities holding true:

$\frac{P(\lambda_1)}{0!}=0,\frac{P'(\lambda_1)}{1!}=0,\ldots,\frac{P^{(m_1-1)}(\lambda_1)}{(m_1-1)!}=0$

$\frac{P(\lambda_2)}{0!}=0,\frac{P'(\lambda_2)}{1!}=0,\ldots,\frac{P^{(m_2-1)}(\lambda_1)}{(m_2-1)!}=0$

$\ldots$

This implies that each of the following polynomials divides $P$: $(x-\lambda_1)^{m_1}$, $(x-\lambda_2)^{m_2}$, $\ldots$ , $(x-\lambda_k)^{m_k}$.

As the $\lambda_i$ are distinct, we get: $P(x)=Q(x)\prod_{p=1}^{k}(x-\lambda_p)^{m_p}$.

As $P$ is of degree less than $n$, this gives $Q=0$, so $P=0$, so all the $a_i$'s are zero.

Hence, the matrix $V_{m_1,\ldots,m_k}(\lambda_1,\ldots,\lambda_k)$ has linearly independent rows; it's invertible.

This is a rather cumbersome proof, where I tried to imitate the inductive approach explained for example here.

Let us first have a look at inductive step for the case of $V_{3,3}(x,y)$

$|V_{3,3}(x,y)|= \begin{vmatrix} 1 & 0 & 0 & 1 & 0 & 0\\ x & 1 & 0 & y & 1 & 0\\ x^2 & 2x & 1 & y^2 & 2y & 1\\ x^3 & 3x^2 & 3x & y^3 & 3y^2 & 3y \\ x^4 & 4x^3 & 6x^2 & y^4 & 4y^3 & 6y^2\\ x^5 & 5x^4 & 10x^3 & y^5 & 5y^4 & 10y^3 \end{vmatrix}\overset{(1)}= \begin{vmatrix} 1 & 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & y-x & 1 & 0\\ 0 & x & 1 & y^2-xy & 2y-x & 1\\ 0 & x^2 & 2x & y^3-xy^2 & 3y^2-2xy & 3y-x \\ 0 & x^3 & 3x^2 & y^4-xy^3 & 4y^3-3xy^2 & 6y^2-3xy\\ 0 & x^4 & 4x^3 & y^5-xy^4 & 5y^4-4xy^3 & 10y^3-6xy^2 \end{vmatrix}\overset{(2)}= \begin{vmatrix} 1 & 0 & (y-x)1 & 1 & 0\\ x & 1 & (y-x)y & 2y-x & 1\\ x^2 & 2x & (y-x)y^2 & 3y^2-2xy & 3y-x \\ x^3 & 3x^2 & (y-x)y^3 & 4y^3-3xy^2 & 6y^2-3xy\\ x^4 & 4x^3 & (y-x)y^4 & 5y^4-4xy^3 & 10y^3-6xy^2 \end{vmatrix}\overset{(3)}= (y-x) \begin{vmatrix} 1 & 0 & 1 & 1 & 0\\ x & 1 & y & 2y-x & 1\\ x^2 & 2x & y^2 & 3y^2-2xy & 3y-x \\ x^3 & 3x^2 & y^3 & 4y^3-3xy^2 & 6y^2-3xy\\ x^4 & 4x^3 & y^4 & 5y^4-4xy^3 & 10y^3-6xy^2 \end{vmatrix}\overset{(4)}= (y-x) \begin{vmatrix} 1 & 0 & 1 & 0 & 0\\ x & 1 & y & y-x & 1\\ x^2 & 2x & y^2 & 2y^2-2xy & 3y-x \\ x^3 & 3x^2 & y^3 & 3y^3-3xy^2 & 6y^2-3xy\\ x^4 & 4x^3 & y^4 & 4y^4-4xy^3 & 10y^3-6xy^2 \end{vmatrix}\overset{(5)}= (y-x)^2 \begin{vmatrix} 1 & 0 & 1 & 0 & 0\\ x & 1 & y & 1 & 1\\ x^2 & 2x & y^2 & 2y & 3y-x \\ x^3 & 3x^2 & y^3 & 3y^2 & 6y^2-3xy\\ x^4 & 4x^3 & y^4 & 4y^3 & 10y^3-6xy^2 \end{vmatrix}\overset{(6)}= (y-x)^2 \begin{vmatrix} 1 & 0 & 1 & 0 & 0\\ x & 1 & y & 1 & 0\\ x^2 & 2x & y^2 & 2y & y-x \\ x^3 & 3x^2 & y^3 & 3y^2 & 3y^2-3xy\\ x^4 & 4x^3 & y^4 & 4y^3 & 6y^3-6xy^2 \end{vmatrix}\overset{(7)}= (y-x)^3 \begin{vmatrix} 1 & 0 & 1 & 0 & 0\\ x & 1 & y & 1 & 0\\ x^2 & 2x & y^2 & 2y & 1 \\ x^3 & 3x^2 & y^3 & 3y^2 & 3y\\ x^4 & 4x^3 & y^4 & 4y^3 & 6y^2 \end{vmatrix}\overset{(8)}= (y-x)^3|V_{2,3}(x,y)|$

(1): Subtracting each row $x$-times from the next row.
(2): Laplace expansion with respect to the first column.
(4): Subtracting the 3-rd column from the 4-th column.
(6): Subtracting the 4-th column from the 5-th column.

Based on the above computations, it seems reasonable to conjecture that $$|V_{m_1,\dots,m_k}(\lambda_1,\dots,\lambda_k)|= \prod_{j=2}^k (\lambda_j-\lambda_1)^{m_j} |V_{m_1-1,\dots,m_k}(\lambda_1,\dots,\lambda_k)| \tag{A}.$$

Using (A) we could already prove the general formula using induction; in the inductive we would get as the term containing $(\lambda_j-\lambda_1)$ the following: $$(\lambda_j-\lambda_1)^{m_j}(\lambda_j-\lambda_1)^{(m_1-1)m_j}=(\lambda_j-\lambda_1)^{m_1m_j}.$$

The proof of (A) seems to be just checking that binomial coefficients appearing in this matrix behave in general in the same way as in the above example, when doing the same operations. It seems to be quite messy to write this down in detail, so I did not do it. I will just say that when subtracting the rows we will get $$\binom{k-1}{l-1}\lambda_1^{k-l}-\lambda_1\binom{k-2}{l-1}\lambda_1^{k-l-1}=\lambda_1^{k-l}\left(\binom{k-1}{l-1}-\binom{k-2}{l-1}\right)= \lambda_1^{k-l}\binom{k-2}{k-2}$$ when working in the first block and $$\binom{k-1}{l-1}\lambda_j^{k-l}-\lambda_1\binom{k-2}{l-1}\lambda_j^{k-l-1}$$ for the block corresponding to $\lambda_j$. It remains to check that in the first block we indeed have the entries from the matrix of the degree $(n-1)$ and that everything works also for column operations.