1. ## Determinant

Let $a_{1},a_{2},...,a_{n}$ in $R$ , where $n \geq 2$ . Show that

$\left| \begin{array} {ccccc}1 & a_{1} & a_{1}^{2} & ... & a_{1}^{n-1} \\ 1 & a_{2} & a_{2}^{2} & ... & a_{2}^{n-1} \\ ... & ... & ... & ... & ... \\ 1 & a_{n} & a_{n}^{2} & ... & a_{n}^{n-1} \end{array} \right|$ = $\prod_{1 \leq i < j \leq n} (a_{j}-a_{i})$.

Suggestion: Do it by induction.

Now clearly this is true for n=2. Also, it can quite easily be shown for n=3.

Now following the protocol for a proof by induction, we assume that it is true for n=k. Now it must somehow be possible to show that it is true for n=k+1 which would complete the proof.

For now all I have is that the matrix for n=k (for which we know the determinant is $\prod_{1 \leq i < j \leq k} (a_{j}-a_{i})$) is the matrix for n=k+1 minus the k+1th row and the k+1th column. Is there any way this can be used to prove the expected result that det(matrix for n=k+1)= $\prod_{1 \leq i < j \leq k+1} (a_{j}-a_{i})$ ?

Thanks!!

2. Hi, writing the matrix of order k+1 and making a development with respect to the last column (the usual way of computing the determinant), you will see exactly where to use the induction hypothesis.

3. Hmmmm ok that's what I did

Let A= $\left| \begin{array} {cccccc}1 & a_{1} & a_{1}^{2} & ... & a_{1}^{k-1} & a_{1}^{k} \\ 1 & a_{2} & a_{2}^{2} & ... & a_{2}^{k-1} & a_{2}^{k} \\ ... & ... & ... & ... & ... \\ 1 & a_{k} & a_{k}^{2} & ... & a_{k}^{k-1} & a_{k}^{k} \\ 1 & a_{k+1} & a_{k+1}^{2} & ... & a_{k+1}^{k-1} & a_{k+1}^{k} \end{array} \right|$

Then by expansion of det(A) by the (k+1)th column $det (A)= a_{1}^{k} (-1)^{k+2} det(A_{1 \hspace {2mm} k+1})$ $+ a_{2}^{k} (-1)^{k+3} det(A_{2 \hspace {2mm} k+1}) + ...$ $- a_{k}^{k} det(A_{k \hspace {2mm} k+1}) + a_{k+1}^{k} \prod_{1 \leq i < j \leq k} (a_{j}-a_{i})$

Now I'm still not sure how to use induction on this?

4. The matrix $A_{i \ k+1}$ in your development have the same structure as the original one, you can apply the induction hypothesis on them as well.