Proving $|V(a_0,...,a_{n-1})|=\Pi_{0\leq i

Question

Note that this question is similar but slightly different than this. If you believe the answer from that question could be applicable, please explain why it still works for a matrix that has been transposed

The matrix $A$ is defined as:
$V=\left(\begin{array} & 1 & a_0 & a_0^2 & \cdots & a_0^{n-1} \\ 1 & a_1 & a_1^2 & \cdots & a_1^{n-1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & a_{n-1} & a_{n-1}^2 & \cdots & a_{n-1}^{n-1} \\\end{array}.\right)$

Such that $a_0,a_1,..,a_{n-1}\in \mathbb{C}$

Prove that:
$|V(a_0,...,a_{n-1})|=\Pi_{0\leq i<j<n-1}(a_j-a_i)$

For example:

$V(3,2,4)=\left( \begin{array} &1&3&9 \\ 1&2&4 \\ 1&4&16 \end{array} \right)$ such that: $\begin{vmatrix} 1&3&9 \\ 1&2&4 \\ 1&4&16 \end{vmatrix}=(4-2)(4-3)(2-3)$

Use the following steps in your proof:
-$C_n-a_0C_{n-1}\rightarrow C_n$
-$C_{n-1}-a_0C_{n-2}\rightarrow C_{n-1}$
-Until $C_2-a_0C_1\rightarrow C_2$
-Use induction/recursion to arrive at a solution

I don't see how your question is different; in particular, the second answer (by user63181) uses exactly the step you mention (up to transposition). — Arnaud D., Jul 12 '17 at 13:24
If you know $\det(A)=\det(A^t)$, you can follow the link in Arnaud D. answer. If not, use the given instruction. I cannot say it better. — Fakemistake, Jul 12 '17 at 13:33
@ArnaudD. can you explain why that works even though the matrix is transposed? — CluelessButCurious, Jul 12 '17 at 13:41
And even if you don't know that $\det(A)=\det(A^t)$ (or even what $A^t$ means), the algebraic manipulations are exactly the same, wether you see your matrix on one side or another. — Arnaud D., Jul 12 '17 at 13:42

score 0 · Answer 1 · answered Jul 12 '17 at 14:37

Most of this answer is just an explanation of the excellent answer given by @user63181 in an extremely similar question. The main difference between this question and the one mentioned is that my matrix is the transposed version of the matrix mentioned.

If $n=3$:

$V=\begin{vmatrix} 1 & a_0 & a_0^{n-1} \\ 1 & a_1 & a_1^{n-1} \\ 1 & a_{n-1} & a_{n-1}^{n-1} \\\end{vmatrix}=\begin{vmatrix} 1 & a_0 & a_0^{n-1} \\ 1-a_0 & a_1-a_0^2 & a_1^{n-1}-a_0^n \\ 1-a_0 & a_{n-1}-a_1a_0 & a_{n-1}^{n-1}-a_1^{n-1}a_0\end{vmatrix}$

$det(V(a_0,...,a_{n-1}))=\begin{vmatrix} a_1-a_0^2 & a_1^{n-1}-a_0^n \\a_{n-1}-a_1a_0 & a_{n-1}^{n-1}-a_1^{n-1}a_0 \end{vmatrix} -a_0\begin{vmatrix} 1-a_0& a_1^{n-1}-a_0^n \\ 1-a_0 & a_{n-1}^{n-1}-a_1^{n-1}a_0 \end{vmatrix} +a^{n-1}\begin{vmatrix} 1-a_0 & a_1-a_0^2 \\1-a_0 & a_{n-1}-a_1a_0 \end{vmatrix}=\left((a_1-a_0^2)(a_{n-1}^{n-1}-a_1^{n-1}a_0)-(a_1^{n-1}-a_0^n)(a_{n-1}-a_1a_0)\right)+(1-a_0)\left(-a_0\left((a_{n-1}^{n-1}-a_1^{n-1}a_0)-(a_1^{n-1}-a_0^n)\right)+a^{n-1}\left((a_{n-1}-a_1a_0)-(a_1-a_0^2)\right)\right)=-(a_0-a_1)(a_0-a_{n-1})(a_1-a_{n-1})=\Pi_{0\leq i<j<n-1}(a_j-a_i)\blacksquare$

Proving $|V(a_0,...,a_{n-1})|=\Pi_{0\leq i

1 Answers1