1. ## Sylvester Matrix

Hello I am in the process of proving that the determinant of the sylvester matrix gives the resultant

f(x) = a_0 x^m + ... + a_{m-1} x + a_m
g(x) = b_0 x^n +... + b_{n-1} x + b_n

(can't get latex to work)

I am supposed to show that the determinant of the Sylvester matrix associated to f,g has degree less than or equal to mn. (see PlanetMath: proof that Sylvester's matrix equals the resultant) I don't understand how it is clear that the degree is less than or equal to mn. I do understand that each a_k or b_k has degree k.

Any help would be appreciated , thank you

2. You have to show that the determinant of the Sylvester matrix associated with f and g has degree less than mn, or m + n?

3. Originally Posted by Ackbeet
You have to show that the determinant of the Sylvester matrix associated with f and g has degree less than mn, or m + n?
It has degree less than $\displaystyle mn$.

Hint : remember that the determinant of the matrix $\displaystyle (a_{ij})$ can be written $\displaystyle \sum_{\sigma \in S_n} (\mbox{sgn }\sigma) a_{1\sigma(1)}\dots a_{n\sigma(n)}$, where the sum is taken over the symmetric group $\displaystyle S_n$ and $\displaystyle \mbox{sgn }$ is the sign.