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 !

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    $\begingroup$ Some examples are in math.uconn.edu/~kconrad/blurbs/linmultialg/univid.pdf (theorems in section 3 and proofs on section 4). $\endgroup$ – KConrad Nov 26 '14 at 1:45
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    $\begingroup$ Thanks KConrad for this well-written document. I really like the " elevator trick " : go up to generic matrices, then down to the complex case, and get back to the original one. $\endgroup$ – Ayman Moussa Nov 26 '14 at 8:27

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 $\mathbb Z$.

  • $\begingroup$ You can't deduce that $Q$ must be 1. Why not $-1$? $\endgroup$ – KConrad Nov 26 '14 at 1:47
  • $\begingroup$ @Conrad. This is a choice. You have to square roots. The convention is that ${\bf Pf}(J_{2n})=1$. $\endgroup$ – Denis Serre Nov 26 '14 at 6:24
  • $\begingroup$ Thanks Denis. Actually, I did search for that kind of stuff in the french edition of your book on matrices, but did not find it. I eventually discovered it in the english version. $\endgroup$ – Ayman Moussa Nov 26 '14 at 8:46
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    $\begingroup$ @DenisSerre: I didn't question your definition of the Pfaffian, but rather only that you can't deduce from what you write that $Q$ must equal 1. Instead, since you show $Q$ divides $P$ you can redefine $Q$ to make it 1. Separately, instead of writing that $P/Q$ is irreducible you meant that $P/Q$ is in lowest terms. $\endgroup$ – KConrad Nov 26 '14 at 13:14
  • $\begingroup$ @Conrad. Yes, the correct terminology is probably in lowest terms. $\endgroup$ – Denis Serre Nov 26 '14 at 14:11

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