Nilpotent operators (Definition 7.6.4) are the building blocks for understanding non-diagonalizable maps. On each generalized eigenspace , the operator is nilpotent, so understanding nilpotent operators is the key to understanding all operators over an algebraically closed field.
7.7.1 Basic Properties
Example 7.7.1 (Basic examples of nilpotent operators).
- The zero operator on is nilpotent with index 1.
- The shift operator on with matrix , which sends and for , is nilpotent with index .
- Any strictly upper-triangular matrix (zeros on and below the diagonal) is nilpotent: maps into , so .
Proposition 7.7.2 (Characterizations of nilpotent operators). For with , the following are equivalent:
- is nilpotent.
- .
- .
If is algebraically closed and , they are also equivalent to:
- is the only eigenvalue of .
Proof. (b) (a) is clear. (a) (c): if , every vector is killed by . (c) (b): by Proposition 7.6.2(b), so .
(a) (d): If and with , then , so . And is an eigenvalue: is not injective, since otherwise would be injective on .
(d) (c): By the generalized eigenspace decomposition (Theorem 7.6.11), over the eigenvalues . If the only eigenvalue is , then . ∎
Note. Over a field that is not algebraically closed, the absence of nonzero eigenvalues does not imply nilpotence: the rotation by of has no eigenvalues at all. What remains true over every field is that is nilpotent if and only if . If is nilpotent, the next corollary gives a strictly upper-triangular matrix for , and then by Corollary 7.5.4(b); the converse is a consequence of the Cayley–Hamilton Theorem (Theorem 7.9.1).
Corollary 7.7.3 (Nilpotent operators have strictly upper-triangular matrices). Let be finite-dimensional. If is nilpotent, then there exists a basis of in which the matrix of is strictly upper triangular. This holds over every field.
Proof. We prove this by induction on . If there is nothing to prove, and if any basis works.
Let and . Consider . We have : otherwise would be surjective, hence bijective (Corollary 6.1.16), and then no power of could be zero on . The subspace is -invariant, and is nilpotent. By induction, has a basis with for each . Extend it to a basis of . Each lies in , which is spanned by vectors preceding . So in this basis every basis vector is mapped into the span of the preceding ones, which means the matrix is strictly upper triangular. ∎
7.7.2 The Nilpotent Basis Construction
The key to understanding nilpotent operators is building a basis adapted to the “chains” of vectors connected by .
Definition 7.7.4 (Chains). Let be nilpotent. A chain (or Jordan chain) of length is a sequence of vectors with such that: The vector is called the head of the chain, and is the tail. Equivalently, the chain is for a vector with and .
Note. In a chain, , , and generally , since while . The chain records how moves vectors toward the kernel.
Lemma 7.7.5 (Key lemma for chain construction). Let be nilpotent and . Suppose is a subspace satisfying . Then:
- is injective.
- .
- .
Proof. (a) If and , then , using .
(b) For , we have , so , meaning .
(c) Let with . Then , so , hence . ∎
Theorem 7.7.6 (Structure theorem for nilpotent operators). Let be finite-dimensional and let be nilpotent with index (so but ). Then there exists a basis of of the form where each sequence is a chain: The chain lengths satisfy and , and the number of chains is .
Proof. We construct subspaces for , working down from , together with bases of , such that with injective on .
First step: Let be a complement of in (extend a basis of to a basis of , and let be the span of the added vectors ). Since , . These vectors will be the heads of the chains of length .
Inductive step: Suppose and have been constructed for some , with . By Lemma 7.7.5 applied to , the map is injective on , so is a linearly independent set of vectors spanning ; moreover and . Hence a basis of together with is linearly independent; extend it by vectors to a basis of . Set Then . The new vectors (possibly none) will be the heads of chains of length .
Assembling the basis: Since , the decompositions give so is a basis of . Put . Each gives the chain of length : its vectors lie in , because , and because it is a basis vector. Every element of either belongs to or is the image under of exactly one element of ; following these images upward, every basis vector lies in exactly one chain. Thus the basis is partitioned into chains.
The longest chains have length (their heads form ), and ordering the chains by decreasing length gives . Each chain has exactly one vector in , its tail, so . ∎
Note. The chain lengths form a partition of . We will see in Proposition 7.7 that this partition depends only on : the choices made in the construction do not affect it.
7.7.3 Jordan Blocks for Nilpotent Operators
The chain basis gives the matrix of a nilpotent operator a very specific form.
Definition 7.7.7 (Nilpotent Jordan block). A nilpotent Jordan block of size is the matrix It has 1’s on the superdiagonal and 0’s elsewhere.
Corollary 7.7.8 (Matrix form of nilpotent operators). With respect to a chain basis ordered as , the matrix of a nilpotent operator is block diagonal:
Proof. Each chain spans a -invariant subspace, and is the direct sum of these subspaces, so the matrix is block diagonal (Theorem 7.3.4). Within the chain , we have for and : the -th column of the block has a single , in row . This is exactly . ∎
Example 7.7.9 (Nilpotent operator with partition ). Consider nilpotent with chain lengths . A chain basis might be: where , , , , , and .
The matrix is:
Proposition 7.7.10 (Uniqueness of the chain lengths). Let be finite-dimensional, nilpotent, and let for . In any basis of consisting of chains for , the number of chains of length is , and the number of chains of length exactly is . In particular, the partition of Theorem 7.7.6 is uniquely determined by .
Proof. Let be a basis consisting of chains of lengths , with and . Then if , and if . The vectors with are distinct basis vectors, hence linearly independent. So vanishes exactly when for all , and Therefore is the number of with . The number of chains of length exactly is the number of length minus the number of length . Since the are determined by , so is the partition. ∎
Note. This dimension-counting argument is the key to proving uniqueness of the Jordan form in general. It also shows that two nilpotent matrices are similar if and only if they have the same chain lengths: similar matrices have kernels of powers of the same dimensions, and two nilpotent matrices with the same chain lengths are both similar to the same block matrix .