This time, I want to share an observation on the Jordan decomposition, which is the main tool needed to show the existence of the Jordan normal form. Let me begin by introducing a more general notation, and show that the Jordan decomposition satisfies a kind of universal property.
Let be a vector space over a field
and
a linear map. We say that a subspace
is
-invariant if
. Another way to interpret this is to consider the
-algebra
; then
is an
-module and the
-submodules of
are exactly the
-invariant subspaces of
.
An -decomposition of
is a decomposition
of
-submodules such that, for every
-submodule
of
, one has
.
Clearly, there always exists a trivial -decomposition of
, namely
itself. One can define a partial order on the set of
-decompositions:
Let and
be two
decompositions. We say that
if, for every
, there exists an
with
.
Clearly, the trivial -decomposition is the maximum with respect to this order. One can ask whether a minimal
-decomposition exists. In case it exists, it has a nice property:
Assume that is a minimal
-decomposition. Let
be another
-decomposition. Then, for every
, there exists a subset
with
.
Define . Then
is a disjoint union. Now,
and
form a direct sum, whence
for all
.
Now assume that for some
; let
. Now, as
, we can write
with
. Moreover, write
with
. Clearly, we must have
for every
. As
we have
for all
, whence
for all
. But this implies
, a contradiction.
Now one can ask when such a decomposition exists, and if it can be computed. An important case in which this is true is the one where is a finitely dimensional vector space over an algebraically closed field
; for example,
and
.
Let . The generalized eigenspace of
with respect to
is
In case , one has that
. Hence, generalized eigenspaces can be efficiently computed. Moreover, we have
, and a simple argument shows that either both are trivial or both are non-trivial. Hence, the
with
are exactly the zeroes of the characteristic polynomial of
.
Now note that . Hence,
is
-invariant. We now have three lemmas:
Let be
different eigenvalues of
. Then
is a direct sum.
Let with
. We have to show that
for all
. Assume that not all
are zero, and that the relation is chosen minimal with respect to the number of nonzero
.
Let with
, and choose
with
. If
, then
yields a second relation with
. By minimality, we must have
for all
.
We will show that is injective for
, which gives
for all
and, therefore,
, a contradiction.
Let with
. Assume that
and let
be maximal with
; in that case,
and
whence we get , which is only possible for
, a contradiction. Hence, we must have
, i.e.
is injective on
.
Assume that and let
. Then there exists an
-invariant subspace
such that
.
Set . ConsiderS the chains
and
Clearly, there exists an with
as
. Now one easily shows
for all
. By the Dimension Formula, we have
for all , whence
for all
. But then
and
.
Set and let
, i.e. let
with
. Then
whence is
-invariant. Now it suffices to show that
. Now
, whence for every
there exists some
with
. But this means that
is surjective, whence
. But then
.
Assume that and that the characteristic polynomial
of
splits into linear factors. Let
be all eigenvalues of
. Then
.
We proceed by induction on . For
this is cleary. Hence, assume
and let
be an eigenvalue of
. Choose an
-invariant subspace
with
. We have
, whence
. Now
whence the characteristic polynomial of splits into linear factors as well.
Let be the eigenvalues of
. Then, by induction, we have
. Now
, whence
.
Finally, note that for all
, as this would contradict
. Therefore,
. Moreover, we must have
as the dimensions of the generalized eigenspaces for all
must be non-zero, whence "
" must hold. The converse holds because every non-trival generalized eigenvector yields a non-trivial eigenvector to the same value.
Therefore, we get:
Let be algebraically closed and assume that
. Then, for every endomorphism
of
, there exist
such that
is an -decomposition.
We have to show that this yields an -decomposition. For that, let
be a
-invariant subspace of
. Consider
; this is an endomorphism of
whose set of eigenvalues is a subset of the set of eigenvalues of
. Hence, by the previous lemma applied to
, we have
what we had to show.
We can now prove our main result, namely that the generalized eigenspace decomposition is exactly the minimal -decomposition of
:
Let be algebraically closed and
. Then the minimal
-decomposition of
is given by
Note that we do not need that is algebraically cloesd, but only that
splits over
.
Let be a
-decomposition. Assume that there exists some
and
such that
with
; if this would not exist, we would have
.
Assume that we can find eigenvectors and
. Then
, whence
is an eigenvector as well. But then
is an
-invariant subspace of
with
, but
, a contradiction that
is a
-decomposition.
We now show that contains an eigenvector. As
is
-invariant, we can consider
. Now
, whence we must have
. Hence, there exists some
,
with
.
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