Generalized Eigenspaces and Jordan Normal Form — Studies in Algebra and Group Theory

11 Generalized Eigenspaces and Jordan Normal Form

26 September 2025 · Denis Auroux

Recap: eigenvectors and triangular form

Last time, we talked about linear operators , their invariant subspaces ( such that ), and eigenvectors ( such that , i.e. ).

Over any field:

eigenvectors need not exist; eigenvectors for distinct are linearly independent;
if there exist distinct eigenvalues, then is diagonalizable: there exists a basis such that .

We saw:

Proposition 11.1. If is algebraically closed (e.g. ) and a finite-dimensional vector space over , then any linear operator has an eigenvector.

Corollary 11.2. Given over an algebraically closed field , there exists a basis of in which the matrix of is upper-triangular, (i.e. ; each subspace is invariant).

Proof. Induction on .

If , then any nonzero vector gets mapped to a multiple of itself ✓ (any matrix is triangular).
Assume the result is true for dimension , and consider with . Since is algebraically closed, has at least one eigenvector , . Let , and let , with the quotient map (i.e. work “mod ”). Since , , so factors through . So there exists such that commutes (intuitively, “ mod ”).

Since , by the induction hypothesis there exists a basis of such that . Let be such that (i.e. ). (Note: span span , hence are a basis.) Then (In fact, in the basis ,

∎

Now suppose we have and a basis of such that is upper-triangular, and denote by the diagonal entries of .

Lemma 11.3. is invertible iff all the diagonal entries of are nonzero.

Proof. (Seen last time.)

If all are nonzero, then is surjective (hence an isomorphism) since etc. for all .
If , then , so has nontrivial kernel (since ); hence , and is not invertible.

∎

Corollary 11.4. The following are equivalent:

is an eigenvalue of ;
is not invertible;
for some diagonal entry of any upper-triangular matrix representing .

((1) (2) since eigenvectors , and (2) (3) by applying the lemma to and the matrix .)

Next goal: further study of invariant subspaces and eigenvalues for linear operators over an algebraically closed , especially : Jordan normal form. (This is Axler ch. 8; we’ll return to the skipped chapters 6 and 7 soon.)

Generalized kernels

First we talk about generalized eigenspaces. Recall .

Definition 11.5. The generalized kernel of is

These are all the vectors that are eventually sent to by repeatedly applying .

Observe:

(since , ...);
if , then the sequence remains constant after that!

(Proof: . So .)

Since the sequence stops increasing after at most steps,

Example 11.6. , , , i.e. . Then , but .

Lemma 11.7. If , then .

Proof. If , then (using ), so . Hence . By the dimension formula, . ∎

Definition 11.8. Say is nilpotent if there exists such that , i.e. .

Generalized eigenspaces

Now we can do the same thing to eigenspaces:

Definition 11.9. is a generalized eigenvector with generalized eigenvalue if , i.e. there exists such that . Call the generalized eigenspace.

Definition 11.10. The multiplicity of the eigenvalue is the dimension of the generalized eigenspace

In a basis where the matrix of is triangular, this is the number of times appears on the diagonal! (This will be clearer later...)

Proposition 11.11 (Invariant splitting). and are invariant subspaces of , and .

Proof.

Let ; then , hence . But commutes with , so this implies , hence .
If , then .
The lemma above, applied to , says .

∎

Proposition 11.12 (Independence of generalized eigenspaces). The subspaces are independent: , for all .

Proof. Assume , , the distinct. We’ll show (same for the others).

If , let be the largest integer such that (but , so ).

Observe: is the sum of and (because the operators commute, and ).

Contradiction; hence , and similarly for all . ∎

Theorem 11.13. If is algebraically closed, a finite-dimensional vector space over , and , then decomposes into the direct sum of the generalized eigenspaces of :

Proof. By induction on ! (The result is clear for .) Assume the result holds up to dimension , and consider the case .

We’ve seen before: algebraically closed has at least one eigenvalue . Let By the invariant splitting proposition above, and are invariant subspaces, and .

Since , induction implies that decomposes into generalized eigenspaces for : where are the eigenvalues of ( eigenvalues of with an eigenvector in ), and Moreover, doesn’t have as an eigenvalue (since ), so .

Now , and Since the generalized eigenspaces contain for all , we find that span ; and they are independent by the independence proposition, hence (and in fact for all ; in other terms, ). ∎

The decomposition gives us bases in which is given by a block diagonal matrix
Moreover, can be represented by a triangular matrix in a suitable basis for (using the theorem seen last time), and since its only eigenvalue is , the diagonal entries are all ! So
We can do more with the blocks , but this requires further study of nilpotent operators (note: is nilpotent!).

Nilpotent operators

Definition 11.14. is nilpotent if there exists such that , i.e. ( for ).

Goal: find a “nice” basis of for a nilpotent operator .

Observe: if , there are 2 cases: either , or but . In the second case, let ; then , so are independent and form a basis, in which .

Jordan’s method generalizes this to higher dimensions:

Proposition 11.15. There exists a basis of in which block diagonal, built from nilpotent Jordan blocks (each basis element the previous one, first basis element ).

Proof. Recall . Assume this is the smallest , i.e. but .

Claim: if a subspace satisfies (), then is injective, , and .

Indeed: for all , , we have and . In particular , i.e. : injective. Also, , and .

First step: let be such that , and pick a basis of [these will yield Jordan blocks of size !]. (E.g. start from a basis of , extend to a basis of by adding vectors , and let be their span.)

Now by the claim, are linearly independent, and their span is contained in but independent of . Start from a basis of , add , and complete to a basis of by adding some other vectors if needed (we could have ). (These will yield blocks of size .) Let . Then .

And so on: given with , take for , and extend by adding vectors as needed to build . This eventually gives a basis of , and rearranging it as we get the result. ∎

Notes not reached in this lecture, covered at the start of Lecture 12

Jordan normal form

We now combine our results to arrive at the Jordan normal form:

Theorem 11.16 (Jordan normal form). Let be a finite-dimensional vector space over an algebraically closed field (e.g. ), and . Then there exists a basis of in which the matrix of is block-diagonal, with each block a Jordan block

Remark.

A size 1 Jordan block is ; size 2: ; ... is diagonalizable all the blocks have size 1.
The values of that appear are exactly the eigenvalues of . There may be several blocks with the same ; their direct sum is the generalized eigenspace .
Proof: we’ve seen (generalized eigenspaces); now is nilpotent, so it can be decomposed into nilpotent Jordan blocks,