Sylvester matrix and GCD degree How to prove that the degree of a $\gcd$ of two polynomials is equal to the dimension of the null space of the Sylvester matrix?
I know that any linear combination of the rows of $S(u,v)$ is a linear combination of the two polynomials, but I don't know how to get the result.
Please any help .
 A: The idea of the answer is given here in the wikipedia entry on the Sylvester matrix of two polynomials. In short: Bézout's identity. But it takes a few more details than those displayed on wikipedia to be convinced. I think...
Let $p, q$ be degree $m,n$ polynomials over a field $F$. And let $S_{p,q}\in M_{m+n}(F)$ denote their Sylvester matrix. 
As you already observed, we can see the transpose $S_{p,q}^T$ as a linear map from $F_{n-1}[X]\times F_{m-1}[X]$ to $F_{m+n-1}[X]$. It suffices to identify $F^k$ with with $F_{k-1}[X]$, the vector space of all polynomials of degree not greater than $k-1$ over $F$, via canonical bases. In particular, this yields 
$$
S_{p,q}^T(x,y)=0\iff px+qy=0
$$
Since $\dim \ker S_{p,q}=\dim \ker S_{p,q}^T$, your question amounts to: how does one prove that 

$$
\dim \ker S_{p,q}^T=\deg \gcd(p,q)
$$

Let $g$ be the gcd of $p$ and $q$, and write $p=gp_0$, $q=gq_0$ with $\gcd(p_0,q_0)=1$. Now observe that for a given pair $(x,y)\in F_{n-1}[X]\times F_{m-1}[X]$
$$
px+qy=0\iff p_0x=-q_0y \iff x=q_0z \;\mbox{and}\; y=-p_0z\quad 
$$
for some unique $z\in F_{d-1}[X]$ where $d$ is the degree of $\gcd(p,q)$.
Therefore we have an isomorphism
$$
F_{d-1}[X]\simeq \ker S_{p,q}^T\quad\mbox{given by}\quad z\longmapsto (q_0z,-p_0z)
$$
The result follows. By rank-nullity, this can also be stated as

$$
\deg \gcd (p,q)=\deg p+\deg q -\mathrm{rank} S_{p,q}
$$

