# 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

• How could that be? The first minor of order 4 is nonzero. Proof for the inverse: Consider some minor being zero. So the rank of the martix is less than $4$. But rank of its submatrix is $4$, since its det is nonzero. Contradiction May 10, 2015 at 10:55
• It is my fault. They should be 0. Please see the new version. Thank you. May 10, 2015 at 10:59

## 1 Answer

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.

• Thank you for a very nice proof. May 10, 2015 at 12:13