Vandermonde determinant and linearly independent Let $a_1,a_2,a_3,b_1,b_2,b_3,b_4,b_5,b_6\in \mathbb{C}$ such that $a_i\not=a_j$ for all $i\not=j.$
If $$\begin{vmatrix}
  a_1 & a_2& a_3 & b_1 \\
  a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
 \end{vmatrix} =0,$$
 $$\begin{vmatrix}
  a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
 \end{vmatrix} =0,$$ 
and 
$$\begin{vmatrix}
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
a_1^6 & a_2^{6} & a_3^{6} & b_6\\
 \end{vmatrix} =0,$$ 
then all minors of order $4$ of the matrix
$$\begin{bmatrix}
  a_1 & a_2& a_3 & b_1 \\
  a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
a_1^6 & a_2^{6} & a_3^{6} & b_6\\
 \end{bmatrix}$$
are $0$. It is stated in a paper that this is true without proof. I believe that it is related with Vandermonde determinant but I do not know how to prove it. Could you please help me or give me an idea? Thank you so much for your help.
Masik
 A: Suppose that none of $a_i$ is zero. Otherwise the proof is obvoius.
Let $x = a_1, y = a_2, z = a_3$. 
$$
\begin{vmatrix}
x^k & y^k & z^k & b_k\\
x^{k+1} & y^{k+1} & z^{k+1} & b_{k+1}\\
x^{k+2} & y^{k+2} & z^{k+2} & b_{k+2}\\
x^{k+3} & y^{k+3} & z^{k+3} & b_{k+3}
\end{vmatrix} = x^k y^k z^k
\begin{vmatrix}
1 & 1 & 1 & b_k\\
x & y & z & b_{k+1}\\
x^2 & y^2 & z^2 & b_{k+2}\\
x^3 & y^3 & z^3 & b_{k+3}
\end{vmatrix} = 0
$$
Since first three columns are linearly independent (they contain nonzero minor $\begin{vmatrix}
1 & 1 & 1\\
x & y & z\\
x^2 & y^2 & z^2\\
\end{vmatrix}$) the last must be their linear combination:
$$
\begin{pmatrix}
b_k \\ b_{k+1} \\ b_{k+2} \\ b_{k +3}
\end{pmatrix} = 
A_k \begin{pmatrix}
1 \\ x \\ x^2 \\ x^3
\end{pmatrix}
+B_k \begin{pmatrix}
1 \\ y \\ y^2 \\ y^3
\end{pmatrix}
+C_k \begin{pmatrix}
1 \\ z \\ z^2 \\ z^3
\end{pmatrix}, \qquad k = 1, 2, 3.
$$
Now, eliminating $A_2, B_2, C_2, A_3, B_3, C_3$:
$$
\begin{pmatrix}
b_{k+1} \\ b_{k+2} \\ b_{k +3}
\end{pmatrix} = 
A_k \begin{pmatrix}
x \\ x^2 \\ x^3
\end{pmatrix}
+B_k \begin{pmatrix}
y \\ y^2 \\ y^3
\end{pmatrix}
+C_k \begin{pmatrix}
z \\ z^2 \\ z^3
\end{pmatrix} = A_{k+1} \begin{pmatrix}
1 \\ x \\ x^2
\end{pmatrix}
+B_{k+1} \begin{pmatrix}
1 \\ y \\ y^2
\end{pmatrix}
+C_{k+1} \begin{pmatrix}
1 \\ z \\ z^2
\end{pmatrix}
\qquad k = 1, 2.
$$
Since the system for $(A_{k+1}, B_{k+1}, C_{k+1})$ is nonsingular, the solution exists and is unique. One solution is obvoius: $A_{k+1} = xA_k, B_{k+1} = yB_{k}, C_{k+1} = yC_k$. Finally
$$
\begin{pmatrix}
b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5 \\ b_6
\end{pmatrix} = 
A_1 \begin{pmatrix}
1 \\ x \\ x^2 \\ x^3 \\ x^4 \\ x^5
\end{pmatrix}
+B_1 \begin{pmatrix}
1 \\ y \\ y^2 \\ y^3 \\ y^4 \\ y^5
\end{pmatrix}
+C_1 \begin{pmatrix}
1 \\ z \\ z^2 \\ z^3 \\ z^4 \\ z^5
\end{pmatrix}
$$
The last column is a linear combination of the first three. No matter which rows we select we always have a zero minor.
