A non-Vandermonde matrix with Vandermonde-like determinant? This question is related to the previous one. Consider $n$ variables $x_1,x_2,\ldots,x_n$ and the following $n\times n$ matrix:
$$
A=\begin{bmatrix}
1 & \cdots &  1 \\ 
x_2 + x_3 + \dots + x_n & \dots &  x_1 + x_2 + \dots + x_{n-1} \\
x_2{x_3}  + x_2{x_4}+ \dots + x_{n-1}x_n & \dots &   x_1{x_2}  + x_1{x_3}+ \dots + x_{n-2}x_{n-1 } \\
\vdots & \dots & \vdots\\
x_2 x_3 \dots x_n & \dots &  x_1 x_2 \dots x_{n-1} \\
\end{bmatrix}.
$$ 
When $i>1$, the element $a_{ij}$ is the sum of all possible products of $i-1$ variables $x_k$'s with distinct indices, except that $x_j$ is not participating in any term on column $j$. Formally,
$$
a_{ij}=\sum_{k_1<\cdots<k_{i-1} \text{ and they are } \ne j} x_{k_1}x_{k_2}\cdots x_{k_{i-1}}.
$$
Of course, when some $x_i=x_j$, $A$ has two equal columns and it becomes singular, but is this the only possibility for $\det A=0$?
 A: The Vee is right. This is a Vandermonde determinant, but I think there is a simpler derivation. To stress the dimension of $A$ and its dependence on $x_1,\ldots,x_n$, we denote the matrix by $A_n(x_1,\ldots,x_n)$ instead. Note that when $i,j>1$, we have
$$
\begin{align*}
a_{ij}-a_{i1}=(x_1-x_j)\sum_{k_1<\cdots<k_{i-2} \text{ and they are } \ne 1,j} x_{k_1}x_{k_2}\cdots x_{k_{i-2}}.
\end{align*}
$$
Therefore, if we subtract the first column from every other column, we get
$$
\begin{bmatrix}
1 & 0\\
\ast & A_{n-1}(x_2,\ldots,x_n)\operatorname{diag}(x_1-x_2,\ldots,x_1-x_n)\ 
\end{bmatrix}
$$
and hence $\det A_n(x_1,\ldots,x_n)=(x_1-x_2)\cdots(x_1-x_n)\det A_{n-1}(x_2,\ldots,x_n)$. Proceed recursively, we obtain
$$
\det A_n(x_1,\ldots,x_n)=\prod_{i<j}(x_i-x_j)
$$
and the determinant vanishes if and only if $x_i=x_j$ for some two $i,j$.
A: The determinant of such a matrix is actually the Vandermonde determinant (up to a sign factor, I think, but I may have made a mistake there). You can find it using the same algorithm:


*

*subtract copies of the first column from all others (no change in determinant)

*subtract multiples of the first row from each other to cancel out $A_{n,1}$ terms (no change in determinant)

*take out the submatrix of second through $n$-th row and column

*factorizing all polynomials, notice that first column is divisible by $(x_2-x_1)$, second column by $(x_3-x_1)$, etc.

*divide each column by its leading element (determinant obtains a prefactor of $\prod_{k=2}^n (x_k - x_1)$)

*find that what's left is exactly of the same form as where you started, just with $x_2, x_3, \ldots, x_n$

*repeat.

The crucial step in doing this formally is writing the difference of two matrix elements in the same row as
$$\begin{aligned}
A_{k,l} - A_{k,m} &= S_l^{k-1} - S_m^{k-1} = (S^{k-1} - x_l S_l^{k-2}) - (S^{k-1} - x_m S_m^{k-2}) = x_m S_m^{k-2} - x_l S_l^{k-2} = \\
&= x_m (x_l S_{m,l}^{k-3} + S_{m,l}^{k-2}) - x_l (x_m S_{m,l}^{k-3} + S_{m,l}^{k-2}) = x_m S_{m,l}^{k-2} - x_l S_{m,l}^{k-2} = (x_m - k_l) S_{m,l}^{k-2}
\end{aligned}$$
where $S^k_{a,b,\ldots}$ is the elementary symmetric polynomial of order $k$ excluding $x_a, x_b, \ldots$.

So to answer your question: the determinant is zero exactly when Vandermonde would be, which in turn is if and only if there are two $x_i$ values that coincide.
A: Consider the polynomial
$$P(x) = (x-x_1) \cdots (x-x_n)$$, and
$$Q_i(x)\colon = \frac{P(x)}{(x-x_i)}$$
for $i=1, \ldots, n$. For instance
$$Q_1(x) = (x-x_2) \cdots (x-x_n) = x^{n-1} - (x_2 + \cdots x_n) x^{n-2} + \cdots + (-1)^{n-1} x_2 \cdots x_{n-1}$$
We have $Q_i(x_k)= 0$ for $k\ne i$, and $Q_i(x_i) = \prod_{k\ne i}(x_i - x_k)$. We conclude that we have the matrix equality
$$ \begin{bmatrix}x_1^{n-1} & x_1^{n-2} & \ldots & 1\\
x_2^{n-1} & x_2^{n-2} & \ldots & 1  \\
\ldots & \ldots & \ldots & 1\\
x_n^{n-1} & x_n^{n-2} & \ldots & 1
 \\ \end{bmatrix} \times \\ \times \begin{bmatrix}
1 & \cdots &  1 \\ 
-(x_2 + x_3 + \dots + x_n) & \dots &  -(x_1 + x_2 + \dots + x_{n-1}) \\
x_2{x_3}  + x_2{x_4}+ \dots + x_{n-1}x_n & \dots &   x_1{x_2}  + x_1{x_3}+ \dots + x_{n-2}x_{n-1 } \\
\vdots & \dots & \vdots\\
(-1)^{n-1}x_2 x_3 \dots x_n & \dots &  (-1)^{n-1}x_1 x_2 \dots x_{n-1} \\
\end{bmatrix}= \\ 
= \begin{bmatrix} (x_1 - x_2) \cdots (x_1 - x_n) & 0 & \ldots & 0 \\
0 & (x_2-x_1) \cdots (x_2-x_n) & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0  & \ldots & (x_n-x_1) \cdots (x_n - x_{n-1}) \end{bmatrix}
$$
Now calculate determinants. Perhaps the signs are a bit off. The first determinant on LHS differs from the standard $V(x_1, \ldots, x_n) = \prod_{j> i}( x_j - x_i)$ by the sign $(-1)^{\binom{n}{2}}$. The second determinant differs from the original one again by the sign $(-1)^{\binom{n}{2}}$. The determinant on RHS is $(-1)^{\binom{n}{2}}\cdot  V^2(x_1, \ldots, x_n)$. We conclude that our determinant equals
$$(-1)^{\binom{n}{2}}\cdot  V(x_1, \ldots, x_n)= \prod_{i < j}(x_i - x_j)$$
Note: the above method provide the inverse of the Vandermonde matrix and is standard.
