How to determine the kernel of a Vandermonde matrix? Given a Vandermonde matrix
$
V= 
\begin{bmatrix} 
1 & 1 & 1 & \ldots & 1 \\\\
x_1 & x_2 & x_3 & \ldots & x_n \\\\
x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\\\
\vdots & \vdots & \vdots & \ddots & \vdots \\\\
x_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \ldots & x_n^{m-1}
\end{bmatrix},
$
when $m=n-1$, $x_i \neq x_j$, what is the kernel of V? I mean when $m=n-1$, the kernel is one-dimensional. Can we present the analytical form for the kernel. 
 A: The answer is pretty much given by darij but it is nice enough (in final form) to spell out a bit further. The short story is that for $n=4$ one vector in (right) kernel is the column vector 
$[\frac{1}{(x_1-x_2)(x_1-x_3)(x_1-x_4)},\frac{1}{(x_2-x_1)(x_2-x_3)(x_2-x_4)},\frac{1}{(x_3-x_1)(x_3-x_2)(x_3-x_4)},\frac{1}{(x_4-x_1)(x_4-x_2)(x_4-x_3)}]^t$ 
and in general one has the  vector whose $jth$ entry is $\frac{1}{\prod_{i \ne j}x_j-x_i}$. Dividing through by the last entry gives 
$[\frac{(x_2-x_4)(x_3-x_4)}{(x_1-x_2)(x_1-x_3)},\frac{(x_1-x_4)(x_3-x_4)}{(x_2-x_1)(x_2-x_3)},\frac{(x_1-x_4)(x_2-x_4)}{(x_3-x_1)(x_3-x_2)},1]^t$
and in general one has the  vector whose $j$th entry is $1$ for $j=n$  and otherwise is is $\prod_{i \ne j,n}\frac{x_i-x_n}{x_j-x_i}$.
The longer story is still fairly compact. Consider the $m \times n$ matrices $V=V_m=V_m(x_1,x_2,\cdots x_n)=$
$$\begin{bmatrix} 
1 & 1 & 1 & \ldots & 1 \\
x_1 & x_2 & x_3 & \ldots & x_n \\
x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \ldots & x_n^{m-1}
\end{bmatrix}$$
The question posed was how to find a vector in the (dimension 1, right) kernel of $V_{n-1}$. If we add any $nth$ row to make an (invertible) matrix $M$, then the last column of $M^{-1}$ will work. The first answer above comes from putting in the final row which makes $V_{n-1}$ into $V_n$. The second comes from using $[0,0,\cdots,0,1]$ as the final row. A related, useful and easier problem is to find an $n$-entry row vector in the (left) kernel of the $n \times (n-1)$ matrix $V_n(x_1,x_2,\cdots,x_{n-1}).$ Try it before reading further. 
If $f(x)=\sum_0^{m-1}a_jX^j$ is a polynomial and $\mathbf{a}=[a_0,a_1,\cdots,a_{m-1}]$ then $\mathbf{a}V =[f(x_1),f(x_2),\cdots,f(x_n)]$. If we want to find the inverse of the square matrix $V_n$ then from $V^{-1}V=I$ we see that the $j$th row of $V^{-1}$ should be the coefficients of the degree $n-1$ polynomial $F$ with $F(x_j)=1$ but $F(x_i)=0$ for $i \ne j$. Evidently, $F(X)=\prod_{i\ne j}\frac{X-x_i}{x_j-x_i}$ since the degree and values are correct. We know how to use the elementary symmetric functions to find the coefficients of $F$ but do not need that knowledge to see that coefficient of $X^{n-1}$ is $1$. Then using $VV^{-1}=I$ we see that the first vector described above is the final column of $V^{-1}$ and is in the kernel of $V_{n-1}$. 
