39vandermonde
thats the clue
Express the Determinant below as a polynomial in alpha's
\begin{bmatrix} 1 & \alpha_1 & \alpha_1^2 & \dots & \alpha_1^{n-1}\\ 1 & \alpha_2 & \alpha_2^2 & \dots & \alpha_2^{n-1}\\ 1 & \alpha_3 & \alpha_3^2 & \dots & \alpha_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 1 & \alpha_m & \alpha_m^2 & \dots & \alpha_m^{n-1}\\ \end{bmatrix}
-
UP 0 DOWN 0 1 9
9 Answers
39is it all that diffiult that it deserves not even a single try in 3 hours??
1vandermonde matrixor vandermonde polynomial
http://en.wikipedia.org/wiki/Vandermonde_matrix
62This one has been unattended since eternity..
hint: substitute a1=a2
(what can you say?)
39we can say that then 1st 2 rows are equal and hence the determinant becomes zero..
so (a1-a2) is a root of the determinant...
so on proceeding...
we find (a2-a3) is a root , (a3-a4) is a root,........................ (am-1-am) is a root...
so determinant can be written as
(a1-a2)(a2-a3)..................................(am-1-am)
62this much is correct bhargav.. but then why will only these terms be there and no other term?
1why can't
(a1-a3)(a1-a4)..(a1-am) (and similar others) be the roots too ??
39ok , so considering all such possiblities,
the determinant can be written as
\prod{(a_{i}-a_{j})}\; where\; 1\leq i<j\leq n