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
.
We have
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
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
.
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