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. 



How to determine the kernel of a Vandermonde matrix?
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 $V\_{n-1}$. 


