Why is a matrix of indeterminates diagonalizable? Fix $n^2$ indeterminates $t_1,\dots, t_{n^2}$. Let $A$ be the algebraic closure of $\mathbb C(t_1,\dots, t_{n^2})$. Consider the $n\times n$ matrix over $A$ whose entries are precisely $t_1, t_2,\dots, t_{n^2}$. Why is this diagonalizable?
I feel I am missing something obvious. My motivation for asking this question comes from a comment on this Mathoverflow answer, which gives a slick proof of the Cayley-Hamilton theorem. I will try to fill in the details; please alert me if there is a gap. If we know this is true, then since the Cayley-Hamilton theorem is easy to verify for diagonalizable operators, we know the matrix $M$ above annihilates its characteristic polynomial. If the characteristic polynomial over $A$ is $p(T)$, then we know that $P(M)=0$. In particular, each entry of $P(M)$ is $0$. But each entry is a polynomial in the indeterminates, which we just saw equals $0$. So no matter which values of $\mathbb C$ we put in for the $t_i$, each entry must vanish. This means every matrix over $\mathbb C$ annihilates its characteristic polynomial.
Maybe there is a more elegant way to finish; my justification seems inadequate, but I can't quite put my finger on why. I would appreciate any comments on this, too. Thank you. 
 A: Let $P(x)$ be the characteristic polynomial of $A$, the $n\times n$ matrix whose entries are the $t_i$'s. Since the matrix is generic, the polynomial must be generic too, and thus it splits in the algebraic closure. In more detail, suppose that it did not split, then some linear factor appears twice. But now fill in particular values for the $t_i$ from $\mathbb C$ for which you know the resulting matrix will not have a repeated root in the characteristic polynomial. Since the $t_i$ are independent over $\mathbb C$, plugging values into these $t_i$ in the polynomial $P(x)$ is the same as plugging these values into the matrix first, and computing the characteristic polynomial of the matrix. In symbols, $P_{A\mid t_i=v_i}(x)=P_A(x)\mid _{t_i=v_i}$. So the assumption that $P(x)$ does not split leads to a contradiction. 
Now, since $P(x)$ splits the matrix $A$ is diagonalizable over the algebraic closure, and this is sufficient for the rest of the proof. 
Note that this nice technique will prove the Cayley-Hamilton over any field, not just $\mathbb C$. The nice thing about this proof is that it is not computational but rather points to the 'true' reason why the claim holds for a general matrix just by the fact that it holds for diagonalizable matrices (for which it is trivial). The trick is to momentarily treat you matrix as a generic one, and only recall you actually had values as entries to begin with, at the end. Luckily, a generic matrix exhibits the most generic behaviour: diagonalizability. 
A similar proof, but more analytic this time, will proceed by slightly perturbing the entires in a given matrix in order to obtain a generic one. 
A: If such a matrix $A$ is not diagonalizable, then its characteristic polynomial $f(x, t_1, \ldots, t_{n^2})$ would have repeated roots.  But then by specializing the $t_i$'s to any particular values of $\mathbb{C}$, this would imply that every matrix over $\mathbb{C}$ has repeated roots.
If you're worried about what happens when you specialize after adjoining roots in the algebraic closure of $\mathbb{C}(\{t_i\})$, you can phrase the argument without leaving the base field.  The polynomial $f(x,t_1,t_2,\ldots,t_{n^2})$ has repeated roots iff the monic greatest common divisor of $f$  and $\partial f/\partial x$ has positive degree.  Then the same will be true of any specialization of $f$.
A: Consider rather $F$ the algebraic closure of $\Bbb Q(t_{ij})$, and some algebraically independent numbers $\pi_{ij}\in\Bbb C$ in such a way that the $\vert\pi_{ii}\vert$ are very large compared to the off diagonal $\pi_{ij}$, and get dramatically larger as $i$ goes from $1$ to $n$. Such a matrix will be diagonalisable over $\Bbb C$ with simple eigenvalues (this is a Gershgorin disk argument).
The caracteristic polynomial $\chi(T)$ of $A$ splits over $F$ so there are $r_1,\dots,r_n\in F$ such that, up to sign, $$\chi(T)=\prod_{i=1,\dots,n} (T-r_i)$$ The field homomorphism sending the $t_{ij}$ to the $\pi_{ij}$ (this is why we chose the $\pi_{ij}$ algebraically independent over $\Bbb Q$) extends to an embedding $F\hookrightarrow\Bbb C$, and by the simplicity of the eigenvalues of the matrix $(\pi_{ij})$ the images of the $r_1,\dots,r_n$ in $\Bbb C$ must be distinct. Thus the caracteristic polynoomial of $A$ has simple roots and $A$ is diagonalisable over $F$.
