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 !



Example of proof using the generic matrix
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 $\mathbb Z$.


