Today, during a lecture, we were posed the question whether , the set of diagonalizable matrices over an algebraically closed field , is Zariski-open, i.e. open in the Zariski topology. This would imply that in case , the set would be open and dense in in the standard (Euclidean) topolgy.
Unfortunately, the answer turns out to be “no” for the case (as well as ):
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 .
Consider the set
Then and is Zariski-open in . In fact, is the complement of a hypersurface in .
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 :
In the standard topology,
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 .
For the proof, we need a little lemma.
Let . Then .
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 .
Let whose characteristic polynomial can be written as , with not necessarily distinct, but monic and irreducible polynomials . There exists a matrix with
clearly, . Moreover, the characteristic polynomial of is given by , i.e. it is squarefree by choice of the . Therefore, and .