# A Note on the Jordan Decomposition.

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 .

Definition.

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:

Definition.

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:

Lemma.

Assume that is a minimal -decomposition. Let be another -decomposition. Then, for every , there exists a subset with .

Proof.

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 .

Definition.

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:

Lemma.

Let be different eigenvalues of . Then is a direct sum.

Proof.

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 .

Lemma.

Assume that and let . Then there exists an -invariant subspace such that .

Proof.

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 .

Lemma.

Assume that and that the characteristic polynomial of splits into linear factors. Let be all eigenvalues of . Then .

Proof.

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:

Corollary (Jordan Decomposition).

Let be algebraically closed and assume that . Then, for every endomorphism of , there exist such that

is an -decomposition.

Proof.

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

We can now prove our main result, namely that the generalized eigenspace decomposition is exactly the minimal -decomposition of :

Theorem.

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 .

Proof.

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 .