Math 414  Fall 1997

The Cyclic Decomposition of a Nilpotent Operator

J.H. Shapiro

1 Introduction.

Suppose T is a linear transformation on a vector space V. Recall Exercise #3 of Chapter 8 of our text, which we restate here as:

1.1 Key Lemma. If, for some vector x in V and some positive integer m, we have:

T^{m−1}x ≠ 0 but T^{m}x = 0

then the list (x, T x, …, T^{m−1}x) is linearly independent.

Suppose in particular that V is finite dimensional with dimension n, and T is nilpotent of index n. Then there is an x ∈ V with T^{n−1}x ≠ 0, and T^{n}x = 0 (the first condition arising from the definition of “index of nilpotence” and the second from the fact that T^{n} is the zero-operator). By the Key Lemma, the list (x, T x, …, T^{n−1}x) forms a basis for V.

Recall that in this case the matrix of T with respect to the basis

(x, T x, …, T^{n−1}x)

has zeros everywhere except on the first subdiagonal, where it has ones. It is more common to write the basis in reverse order, in which case the matrix of T has zeros everywhere but on the first superdiagonal, where it has ones. We will say more about such matrices later.

1.2 Cyclic terminology. Let us call a linear transformation T ∈ L(V) a cyclic operator if there is a vector x ∈ V for which the list (x, T x, …, T^{n−1}x) spans (and is therefore a basis) for V. We call x a cyclic vector for T, and sometimes, when overcome by exuberance, we say that the space V is cyclic for T, or just T -cyclic. If x is a cyclic vector for T we call the corresponding basis (x, T x, …, T^{n−1}x) a cyclic basis for V.

Using this terminology we can restate the Key Lemma like this:

1.3 Proposition. If V is a finite dimensional vector space and T a nilpotent operator with index of nilpotence equal to dim V, then T is cyclic.

1.4 Exercise. (a) Suppose T is a linear transformation of a finite dimensional vector space V, and that x is a cyclic vector for T.

(i) What is the matrix of T with respect to the corresponding cyclic basis of V?

(ii) In terms of this matrix, what is the minimal polynomial of T?

(b) Give an example of a cyclic linear transformation that is not nilpotent.

(c) Give an example of a nilpotent linear transformation that is not cyclic. What’s the most extreme example of this situation?

1.5 The Fundamental Question. What can be said of T if it is nilpotent with index of nilpotence < dim V?

To get a handle on this question, suppose n = dim V (finite) and m is the index of nilpotence of T, with m < n. By the definition of “index of nilpotence” we know that there exists a vector x_{1} ∈ V such that T^{m−1}x_{1} ≠ 0. Since T^{m} = 0, the Key Lemma guarantees that the list (x_{1}, T x_{1}, …, T^{m−1}x_{1}) is linearly independent, and clearly its linear span V_{1} is T -invariant. We are going to prove that:

(*) V_{1} has a T -invariant complement,

i.e. that there exists a T -invariant subspace W_{1} of V such that V = V_{1} ⊕ W_{1}. Once this is done, we’ll be able to restrict T to W_{1}, where it’s still nilpotent of some index m_{1} ≤ m, and find a vector x_{2} ∈ W_{1} such that T^{m_{1}−1}x_{2} ≠ 0, whereupon, as above, the list (x_{2}, T x_{2}, …, T^{m_{1}−1}x_{2}) is linearly independent, and its linear span V_{2} is T - invariant. Upon applying (*) to the restriction of T to W_{1} we obtain a splitting of W_{1} into two subspaces, V_{2} = span (x_{2}, T x_{2}, …, T^{m−1}x_{2}) and a T -invariant complement W_{2}.

Thus far V = V_{1} ⊕ V_{2} ⊕ W_{2}, where all three subspaces are T -invariant, and the restrictions of T to the first two are cyclic. Since V is finite dimensional, we arrive, after a finite number of repetitions of this argument, at a decomposition of V into a finite direct sum of invariant subspaces, with T restricted to each subspace being cyclic.

In summary: once we have established (*), we will have established that V splits into the direct sum of finitely many T -invariant subspaces with the property that the restriction of T to each is cyclic. Put more succinctly, we will have proved:

1.6 The Cyclic Nilpotent Theorem. Every nilpotent linear transformation of a finite dimensional vector space splits into a direct sum of cyclic nilpotent transformations.

We are also interested in the matrix interpretation of this result. It asserts that if T is nilpotent then V has a basis with respect to which the matrix of T is block diagonal, each block being zero except for ones on the first superdiagonal.

If, instead, it is T − λI that is nilpotent for some scalar λ, then each of these cyclic subspaces is T -invariant, and the corresponding matrices have λ down the main diagonal and ones down the first superdiagonal. Such matrices are called λ-Jordan matrices. These observations, along with the Fundamental Decomposition Theorem and the Cyclic Nilpotent Theorem combine to establish:

1.7 Jordan’s Theorem. Suppose T is any linear transformation on a finite dimensional complex vector space V, and suppose λ_{1}, …, λ_{m} are the distinct eigenvalues of T. Then V has a basis with respect to which the matrix of T is block diagonal, where each block is a Jordan λ_{j}-matrix, and every eigenvalue λ_{j} is represented by at least one such block.

The matrix produced in the last theorem is called the Jordan canonical matrix for T. Up to the order in which the Jordan λ_{j}-blocks occur, it is uniquely determined by T.

If T is the left-multiplication operator on F^{n} associated with an n by n F-matrix A, then the Jordan canonical matrix for T is called the Jordan canonical form of A.

2 Toward the proof of (*).

Here, in detail, is the statement of the result we need to prove.

2.1 The Nilpotent-Splitting Theorem. Suppose V is a real or complex vector space (not necessarily finite dimensional), and T ∈ L(V) is nilpotent of index m. Let x be a vector in V with T^{m−1}x ≠ 0. Let V_{1} be the span of the list (x, T x, …, T^{m−1}x) (so V_{1} is T -invariant). Then there is a subspace W_{1} of V that is T -invariant, such that V = V_{1} ⊕ W_{1}.

We will get serious about the proof of this result in the next section. The task will be made considerably easier if we first negotiate some easy preliminaries about inverse images. For the rest of this section V is any vector space and T ∈ L(V). We begin with a familiar definition.

2.2 Definition. If S is a subset of V, then

T^{−1}(S) = {x ∈ V : T x ∈ S}.

In words, T^{−1}(S) is the set of vectors that T sends into S.

We will need some special properties of the inverse image of T when it acts on subspaces of V. We begin with a simple exercise.

2.3 Exercise. Suppose W is a subspace of V. Then

(a) T^{−1}(W) is a subspace of V that contains null T.

(b) T (T^{−1}(W)) = W. (Does this require that W be a subspace?)

(c) W is T -invariant if and only if W ⊂ T^{−1}(W).

2.4 Proposition. T^{−1}(T (W)) = W + null (T).

Proof. T (W) is a subspace of V, so by part (a) of the last Exercise, so is T^{−1}(T (W)). Clearly T^{−1}(T (W)) contains W, and by part (a) of Exercise 2.3 it also contains null T. Because it is a subspace, it therefore contains W + null T.

Conversely, suppose x ∈ T^{−1}(T (W)). then T x ∈ T (W), so T x = T w for some w ∈ W. By linearity, T (x − w) = 0, i.e. x − w ∈ null T, so

x = w + (x − w) ∈ W + null T,

which completes the proof. ///

2.5 Proposition. If W is a T -invariant subspace of V, then so is T^{−1}(W).

Proof. This follows from the containments:

T (T^{−1}(W)) = W ⊂ W + null T ⊂ T^{−1}(W),

in which the first is just part (b) of Exercise 2.3, the second is obvious, and the third follows from parts (a) and (c) of Exercise 2.3. ///

3 Proof of the Nilpotent Splitting Theorem.

The situation: We are given a nilpotent linear transformation T on a vector space V, which need not be finite dimensional. Since m is the index of nilpotence of T, there exists a vector x_{1} ∈ V such that T^{m−1}x_{1} ≠ 0. We know from the Key Lemma (Lemma 1.1 that the list (x_{1}, …, T^{m−1}x_{1}) is linearly independent, and that its linear span V_{1} is T -invariant.

To find: A T -invariant subspace W_{1} of V such that V = V_{1} ⊕ W_{1}.

The proof proceeds by induction on m, the index of nilpotence.

The case m = 1. In this case T is the zero-operator on V, so x_{1} is any non-zero vector in V, V_{1} = span (x_{1}), and we can take W_{1} to be any subspace of V that is complementary to V_{1} (every subspace of V is invariant for the zero-operator).

The Induction Hypothesis. Suppose m > 1 and suppose the result is true for all operators that are nilpotent of index m − 1. The proof now proceeds in four steps.

Step I: Pushing Down. We focus on ran T, which you can easily check is an invariant subspace for T on which T is nilpotent of index m−1, and which is clearly the span of the linearly independent list (T x_{1}, …, T^{m−1}x_{1}) = (y_{1}, …, T^{m−2}y_{1}), where y_{1} = T x_{1}. So we may apply the induction hypothesis to the restriction of T to ran T, with V_{1} replaced by the subspace

Y_{1} = span (y_{1}, …, T^{m−2}y_{1}).  (1)

The result is a T -invariant subspace Y_{2} of ran T such that

ran T = Y_{1} ⊕ Y_{2}  (2)

Step II. Pulling Back. We claim that:

V = V_{1} + T^{−1}(Y_{2}).  (3)

To prove this we write equation (2) as T (V) = T (V_{1}) ⊕ Y_{2}, and distribute the inverse image of T over the sum (Exercise: prove that this is legal) to get:

V = T^{−1}(T (V_{1})) + T^{−1}(Y_{2}).

Now by Proposition 2.4 we have T^{−1}(T (V_{1})) = V_{1} + null T, hence

V = V_{1} + null T + T^{−1}(Y_{2}).

Since null T is a subspace of T^{−1}(Y_{2}), the right-hand side of the last equation is just V_{1} + T^{−1}(Y_{2}), which proves (3).

Equation (3) is a step in the right direction, since it splits V into the sum of V_{1} and the subspace T^{−1}(Y_{2}) which is also T -invariant (by the T -invariance of Y_{2} and Proposition 2.5). Unfortunately T^{−1}(Y_{2}) may have nontrivial intersection with V_{1}, so it is not, in general, a complement for V_{1}. The rest of the argument seeks to remedy this deficiency by “cutting out the excess” from T^{−1}(Y_{2}).

Step III: An important trivial intersection. We claim that, even though V_{1} and T^{−1}(Y_{2}) may have nontrivial intersection, there is still this bit of good news:

V_{1} ∩ Y_{2} = {0}.  (4)

To prove this, note that:

T (V_{1} ∩ Y_{2}) = T (V_{1}) ∩ T (Y_{2})
⊂ Y_{1} ∩ Y_{2}
= {0},

where the first equality is a general property of mappings (can you prove it?), and the second follows from the fact that T (V_{1}) = Y_{1} (definition of Y_{1}) and the T -invariance of Y_{2}. This shows that V_{1} ∩ Y_{2} ⊂ null T, or more precisely:

V_{1} ∩ Y_{2} ⊂ null T ∩ V_{1}  (5)

Now recall the definition of V_{1}:

V_{1} = span (x_{1}, …, T^{m−1}x_{1})

where T^{m−1}x_{1} ≠ 0, but T^{m}x_{1} = 0 (recall that T^{m} is the zero-operator). From this it’s immediate that the only vectors in V_{1} that T annihilates are the scalar multiples of T^{m−1}x, in other words:

null (T) ∩ V_{1} = span (T^{m−1}x) = span (T^{m−2}y) ⊂ Y_{1},

where the last containment follows from our assumption that m > 1.

Summarizing:

V_{1} ∩ Y_{2} ⊂ null T ∩ V_{1} ⊂ Y_{1},

hence

V_{1} ∩ Y_{2} ⊂ Y_{1} ∩ Y_{2} = {0},

as promised.

Step IV: Excising the excess. Let’s summarize what we have so far. We began with a T -invariant subspace V_{1} on which our nilpotent operator T is cyclic, and have:

(a) Produced a T -invariant subspace Y_{2} of T (V) that is complementary to Y_{1} = T (V_{1}).

(b) Proved that V = V_{1} + T^{−1}(Y_{2}), where T^{−1}(Y_{2}) is T -invariant, but may unfortunately intersect V_{1}.

(c) Proved that nevertheless V_{1} ∩ Y_{2} = {0}.

Recall that Y_{2} ⊂ T^{−1}(Y_{2}) (just a restatement of the T -invariance of Y_{2}). By (c) above, V_{1} ∩ T^{−1}(Y_{2}) intersects Y_{2} in the zero-subspace, so if we let Z be any complement of the sum of these two subspaces in T^{−1}(Y_{2}) we have

T^{−1}(Y_{2}) = Y_{2} ⊕ (V_{1} ∩ T^{−1}(Y_{2})) ⊕ Z.  (6)

Thus:

V = V_{1} + T^{−1}(Y_{2}),  restating (b) above
= V_{1} + Y_{2} + (V_{1} ∩ T^{−1}(Y_{2})) + Z,  by (6)
= V_{1} + Y_{2} + Z,  since V_{1} ∩ T^{−1}(Y_{2}) ⊂ V_{1}.

Now by its definition, Y_{2} + Z (= Y_{2} ⊕ Z) lies in T^{−1}(Y_{2}), and it has only trivial intersection with V_{1} ∩ T^{−1}(Y_{2}). Thus it has only trivial intersection with V_{1}, hence V = V_{1} ⊕ Y_{2} ⊕ Z.

We claim that V_{2} = Y_{2} ⊕ Z, is the subspace we seek. We have just shown that V = V_{1} ⊕ V_{2}, so it only remains to show that V_{2} is T -invariant. But this is easy: we already know Y_{2} is T -invariant, and although our choice of Z might have seemed arbitrary, it is a subspace of T^{−1}(Y_{2}) so its image under T lies in Y_{2}. Thus T (V_{2}) ⊂ Y_{2} ⊂ V_{2}, which finishes the proof. ///