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?

 A: 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.
