A problem about Vandemonde matrix

Question

I have this determinant: $A=\begin{vmatrix} 1 & a_{1} & ... & a_{1}^{n-2} & a_{1}^{n} \\ 1 & a_{2} & ... & a_{2}^{n-2} & a_{2}^{n}\\ .& . & ... & . & .\\ .& . & ... & . & .\\ 1 & a_{n} & ... & a_{n}^{n-2} & a_{n}^{n} \end{vmatrix}$

Prove that: $A=\left ( a_{1}+a_{2}+...+a_{n} \right )\det\left ( B \right ),$ inside $B$ is the Vandermonde matrix.

Actually, I have not yet thought of a solution. Hope everyone helps me with the solution of this problem.

Answer 1

Here is an idea to prove the claim. It is based on the Gaussian elimination technique to compute the determinant of the standard Vandermonde matrix. There the first row is subtracted from all the others, then the factors $x_i-x_1$ are factored out. Finally, a column elimination is done to remove any lower order terms from the matrix columns. The remaining submatrix is again a Vandermonde matrix of lower order.

Now if we transfer this to the modified matrix in the question, then the last column of the transformed matrix after one step (subtracting first row from all others, factoring out $x_i-x_1$) consists of entries $x_i^{n-1} + x_i^{n-2}x_1$, $2\le i\le n$. (For the standard Vandermonde these entries would be of the form $x_i^{n-2}$.)

My claim is that after the $k$-step, the last column's entries are of the from $x_i^{n-k} + x_i^{n-k-1}\sum_{j=1}^kx_j$, $k+1\le i\le n$.

Thus after $n-1$ steps, the last remaining entry is $x_n + \sum_{j=1}^{n-1}x_j$ (instead of $1$ for the standard matrix). Now collecting all the factors, we find that the determinant is $\sum_{j=1}^{n}x_j$-times the determinant of the Vandermonde matrix.

Answer 2

It is not an answer but a sketch, suppose the statement true for $n-1$, $n\ge 4$. Take the determinant of the given $n\times n$ matrix , expand from the $n^{th}$ column, that of power $n$. You obtain $\sum a_i^n(\pm M_i)$, where $M_i$ are minors of size $n-1$ now if you take that to be equal the right hand side i.e $(\sum a_i) \det (V)$ then the extra terms sum should be zero, and that extra term(s) is the determinant of the same initial matrix (order $n$) except that the last column is replaced by a minus one exponent and the $n-1$ column has a plus one exponent (by the induction claim, meaning that it has the last two columns equal so its determinant is zero).