Unfortunately, the answer turns out to be “no” for the case (as well as ):

Example.

Let . Consider the matrix as well as the sequence Clearly, . Assume that is open in ; then we must have for almost all . But is in Jordan canonical form, and clearly not diagonalizable; but this means that for all . Therefore, is not open in .

But nonetheless, contains a Zariski-open subset of in case is algebraically closed (which implies that lies dense in ). For that recall that is the characteristic polynomial of .

Proposition.

Consider the set Then and is Zariski-open in . In fact, is the complement of a hypersurface in .

Proof.

Note that in case is squarefree, splits into distinct linear factors since is algebraically closed. Hence, has distinct eigenvalues in and therefore one obtains linearly independent eigenvectors of ; i.e., is diagonalizable over . Therefore, .
Now we show that is a hypersurface in , i.e. there exists a polynomial such that . For that, consider the maps defined by for all . Obviously, these must be polynomials. Next, consider the discriminant of ; this is a polynomial expression in the coefficients of , i.e. in , whose value is zero if, and only if, is squarefree. Therefore, . Finally, is a polynomial, whence is Zariski-open in .

□

Note that the situation is different over :

Proposition.

In the standard topology,

Proof.

Assume that has at least one eigenvalue with imaginary part . If is a sequence of matrices with , each must have an eigenvalue with . But then, for infinitely many , we must have (since is open), whence we cannot have for infinitely many . Hence, is not in the closure of .
Now assume that has only real eigenvalues. Then there exist some with in Jordan canonical form. By pertubing the diagonal elements of slightly, we can obtain a sequence of matrices with . But then, and for every .
Note that this implies ; moreover, this also implies . Hence, the first two equalities hold. The third equality is standard.

□

Also note that for , as the example (which is diagonalizable over , with eigenvalues ) shows. So what about ? In fact, as in the case of , it turns out that is .

Proposition.

We have

For the proof, we need a little lemma.

Lemma.

Let . Then .

Proof.

Let be an arbitrary polynomial. Write , where and is irreducible and monic, . Now the coefficients of all ‘s (except the highest coefficients) are a finite set in of elements, whence there exists sequences with pairwise distinct such that converges to one coefficent of one . In particular, we can construct monic polynomials with , and for every . Even more, we can make sure that every is irreducible; this enforces that is squarefree, i.e. . Therefore, we found a sequence in converging to , whence .

□

Proof (Proof of the Proposition).

Let whose characteristic polynomial can be written as , with not necessarily distinct, but monic and irreducible polynomials . There exists a matrix with where is the companion matrix of ; this is a Frobenius normal form of . Now we can find a sequence of squarefree polynomials such that for every , are pairwise coprime, and that . Then set clearly, . Moreover, the characteristic polynomial of is given by , i.e. it is squarefree by choice of the . Therefore, and .

□