TITLE: Why is a matrix of indeterminates diagonalizable?
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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 $F$.