# Expressing the determinant as the product of two determinants

1. Jun 4, 2012

### Sdelange

This is what the symbols in the question represent( sorry about the syntax) ;
sr = s subscript r
a^r = alpha to the power of r
b^r = beta to the power of r
g^r = gamma to the power of r

Question:

If sr = a^r + b^r + c^r, by expressing the determinant as the product of two determinants, show that

l 3 s1 s2 l
l s1 s2 s3 l
l s2 s3 s4 l

= (a - b)^2 (b - g)^2 (g - a)^2

What I have used so far is the theory behind the product of two matrices being equal to the composition of the matrix corresponding to the linear transformations.

In R2, if you have a set of points x and y, its very simple to work out Ta : (x,y) -> (u,v) and then Tb : (u,v) -> (w,z) . But I'm having problems with actually finding a matrix to work with. Have tried out quite a few things, but I get lost in a whole mess of polynomials. Just need a head start please.

2. Jun 4, 2012

### Vargo

Consider this a hint

$$\left(\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right)$$.

Last edited by a moderator: Jun 4, 2012
3. Jun 4, 2012

### Ray Vickson

I am going to use LaTeX in this posting, and will use the symbols a, b, c instead of $\alpha, \beta, \gamma$. If you introduce the 3-vectors
$$u_0 = (1,1,1), \; u_1 = (a,b,c), \; u_2 = (a^2, b^2, c^2),$$
and define the inner product of two 3-vectors $$x = (x_1,x_2,x_3) \text{ and } y = (y_1,y_2,y_3)$$ as $\langle x , y \rangle= x_1 y_1 + x_2 y_2 + x_3 y_3,$ then your determinant has the form
$$D = \left| \begin{array}{ccc} \langle u_0, u_0 \rangle & \langle u_0,u_1 \rangle & \langle u_0,u_2 \rangle\\ \langle u_1, u_0 \rangle & \langle u_1,u_1 \rangle & \langle u_1,u_2 \rangle\\ \langle u_2, u_0 \rangle & \langle u_2,u_1 \rangle & \langle u_2,u_2 \rangle \end{array} \right|$$

RGV