TITLE: Example of proof using the generic matrix
QUESTION [7 upvotes]: There is a really nice proof of the Cayley-Hamilton Theorem using the generic matrix. I expose it briefly. 
One defines the generic matrix $G:=(X_{ij})_{ij} \in\mathcal{M}_n(\mathbb{Z}[X_{ij}]_{ij})$. 
The discriminant $\Delta_G$ of $\chi_G$ (characteristic polynomial of $G$) is an element of $\mathbb{Z}[X_{ij}]_{ij}$. 
Now $\Delta_G = 0$ is impossible, since it would imply in particular (after specialization) that any element of $\mathcal{M}_n(\mathbb{C})$ has at least one double eigenvalue (which is clearly false).
Whence $\Delta_G \neq 0$. This means that $\chi_G$ has only simple roots (in some field of decomposition, say $\mathbb{K}$), which in turn means that $G$ is diagonalizable in $\mathcal{M}_n(\mathbb{K})$ and $\chi_G(G)$ follows easily. 
We get then the Cayley-Hamilton for any matrix after specialization.
I am searching for other nice (and easy) use of this generic matrix. In fact any generic argument like this one, could be nice also.
I have to precise that I am not an algebraist, so I would really appreciate a simple example. 
I find this proof a bit magic but at the same time very natural (meaning, after all, that $\chi_G(G)=0$ is just a general algebraic identity, just like $(a+b)^2 = a^2+ ab +ba + b^2$, and it is a way to compute it).
Thanks for your help !

REPLY [9 votes]: A famous example is the existence and uniqueness of a polynomial ${\bf Pf}$ in the entries of a $2n\times2n$ alternate matrix, called the Pfaffian, such that 
$${\bf Pf}(A)^2=\det A,\quad\forall A\in{\rm Alt}_{2n},\qquad{\bf Pf}(J_{2n})=1,\quad J_{2n}:=\begin{pmatrix} 0_n & -I_n \\\\ I_n & 0_n\end{pmatrix}.$$
Proof: we know that the determinant of a $2n\times2n$ alternate matrix $A$ with entries in a field $k$ is a square, because $A=P^TJ_{2n}P$ for some $P$. Apply this to the matrix whose entries are indeterminates ; here $k$ is the field of rational fraction ${\mathbb Q}(X)$. Therefore the Pfaffian exists, at least as a fraction $\frac PQ$. We may suppose $P/Q$ irreducible. Then write $P^2=Q^2{\rm Det}$, where ${\rm Det}$ is a polynomial. Every irreducible factor of $Q$ divides $P^2$, hence $P$, and must be constant. Therefore $Q=1$ and ${\bf Pf}=P$ is a polynomial. The same argument gives that it has entries in $\mathbb Z$.