# Vandermonde determinant and linearly independent (corrected version)

This is a corrected version.

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

• I'm curious. What paper are you reading? May 10, 2015 at 11:20

Let \begin{eqnarray}v_i=(a_1^i, a_2^i, a_3^i, b_i)\end{eqnarray} That the three determinants are 0 implies that the row vectors are linearly dependent. In particular, $\text{dim Span}\{v_1, v_2, v_3, v_4\}\leq 3$, $\text{dim Span}\{v_2, v_3, v_4, v_5\}\leq 3$, $\text{dim Span}\{v_3, v_4, v_5, v_6\}\leq 3$. So \begin{eqnarray}\text{dim Span}\{v_1, v_2, v_3, v_4, v_5\}\leq \text{dim Span}\{v_1, v_2, v_3, v_4\}+\text{dim Span}\{v_2, v_3, v_4, v_5\}-3\leq 3\end{eqnarray} Similarly, \begin{eqnarray}\text{dim Span}\{v_1, v_2, v_3, v_4, v_5, v_6\}\leq\text{dim Span}\{v_1, v_2, v_3, v_4, v_5\}+\text{dim Span}\{v_3, v_4, v_5, v_6\}-3\leq 3\end{eqnarray}
So any 4 row vectors are linearly dependent and the determinant of any $4\times 4$ minor is 0.
• Thank you so much for very nice proof. I think you need Vandermonde determinant to show that $Span\{v_2,v_3,v_4\}=3.$ May 11, 2015 at 6:18
• In my proof I didn't use $\text{dim Span}\{v_2, v_3, v_4\}=3$, which is not true in general. Let me spell out more detail of the derivation of the first inequality. Let $V=\text{Span}\{v_1, v_2, v_3, v_4\}$ and $W=\text{Span}\{v_2, v_3, v_4, v_5\}$. Then $V+W=\text{Span}\{v_1, \cdots, v_5\}$, $\text{Span}\{v_2, v_3, v_4\}\leq V\cap W$. Note that \begin{align*}\text{dim}V+W&=\text{dim}V+\text{dim} W-\text{dim}V\cap W\\ \leq& 3+3-\text{dim Span}\{v_2, v_3, v_4\}\\ \leq& 3+3-3\\ =&3\end{align*} May 11, 2015 at 7:58
• Actually, you used it because you need $\dim Span\{v_2,v_3,v_4\}\geq 3,$ which implies $-\dim Span\{v_2,v_3,v_4\}\leq -3$. So, you need to flip the $\geq$ sign. May 11, 2015 at 8:58
• Thanks Masih for the clarification. Indeed I need $\text{dim Span}\{v_2, v_3, v_4\}=3$ which follows from Vandermonde determinant. May 11, 2015 at 9:18