Nilpotence

As an aid to getting a representation theorem more informative than the triangular one, we proceed to introduce and to study a very special but useful class of transformations. A linear transformation is called nilpotent if there exists a strictly positive integer such that ; the least such integer is the index of nilpotence.

Theorem 1. If is a nilpotent linear transformation of index on a finite-dimensional vector space , and if is a vector for which , then the vectors are linearly independent. If is the subspace spanned by these vectors, then there exists a subspace such that and such that the pair reduces .

Proof. To prove the asserted linear independence, suppose that , and let be the least index such that . (We do not exclude the possibility .) Dividing through by and changing the notation in an obvious way, we obtain a relation of the form

It follows from the definition of that

since this contradicts the choice of , we must have for each .

It is clear that is invariant under ; to construct we go by induction on the index of nilpotence. If , the result is trivial; we now assume the theorem for . The range of is a subspace that is invariant under ; restricted to the linear transformation is nilpotent of index . We write and ; then is spanned by the linearly independent vectors . The induction hypothesis may be applied, and we may conclude that is the direct sum of and some other invariant subspace .

We write for the set of all vectors such that is in ; it is clear that is a subspace. The temptation is great to set and to attempt to prove that has the desired properties. Unfortunately this need not be true; and need not be disjoint. (It is true, but we shall not use the fact, that the intersection of and is contained in the null-space of .) That, in spite of this, is useful is caused by the fact that . To prove this, observe that is in for every , and, consequently, with in and in . The general element of is a linear combination of ; hence we have

where is in . It follows that , or , so that is in . This means that is in , so that is the sum of an element (namely ) of and an element (namely ) of .

As far as disjointness is concerned, we can say at least that . To prove this, suppose that is in , and observe first that is in (since is in ). Since is also invariant under , the vector belongs to along with , so that . From this we infer that is in . (Since is in , we have ; and therefore ; from the linear independence of the it follows that , so that .) We have proved that if belongs to , then it belongs also to , and hence that .

The situation now is this: and together span , and contains the two disjoint subspaces and . If we let be any complement of in , that is, if then we may write ; we assert that this has the desired properties. In the first place, and is disjoint from ; it follows that . In the second place, contains both and , so that . Finally, is invariant under , since the fact that implies that . The proof of the theorem is complete. ◻

Later we shall need the following remark. If is any other vector for which , if is the subspace spanned by the vectors , and if, finally, is any subspace that together with reduces , then the behavior of on and is the same as its behavior on and respectively. (In other words, in spite of the apparent non-uniqueness in the statement of Theorem 1, everything is in fact uniquely determined up to isomorphisms.) The truth of this remark follows from the fact that the index of nilpotence of on ( , say) is the same as the index of nilpotence of on ( , say). This fact, in turn, is proved as follows. Since and also (these results depend on the invariance of all the subspaces involved), it follows that the dimensions of the right sides of these equations may be equated, and hence that .

Using Theorem 1 we can find a complete geometric characterization of nilpotent transformations.

Theorem 2. If is a nilpotent linear transformation of index on a finite-dimensional vector space , then there exist positive integers and vectors such that (i) , (ii) the vectors

form a basis for , and (iii) . The integers form a complete set of isomorphism invariants of . If, in other words, is any other nilpotent linear transformation on a finite-dimensional vector space , then a necessary and sufficient condition that there exist an isomorphism between and such that is that the integers attached to be the same as the ones attached to .

Proof. We write and we choose to be any vector for which . The subspace spanned by is invariant under , and, by Theorem 1 , possesses an invariant complement, which, naturally, has strictly lower dimension than . On this complementary subspace is nilpotent of index , say; we apply the same reduction procedure to this subspace (beginning with a vector for which ).

We continue thus by induction till we exhaust the space. This proves the existential part of the theorem; the remaining part follows from the uniqueness (up to isomorphisms) of the decomposition given by Theorem 1. ◻

With respect to the basis described in Theorem 2, the matrix of takes on a particularly simple form. Every matrix element not on the diagonal just below the main diagonal vanishes (that is, implies ), and the elements below the main diagonal begin (at top) with a string of ’s followed by a single , then go on with another string of ’s followed by a , and continue so on to the end, with the lengths of the strings of ’s monotonely decreasing (or, at any rate, non-increasing).

Observe that our standing assumption about the algebraic closure of the field of scalars was not used in this section.

EXERCISES

Exercise 1. Does there exist a nilpotent transformation of index on a -dimensional space?

Exercise 2.

Prove that a nilpotent linear transformation on a finite-dimensional vector space has trace zero.
Prove that if and are linear transformations (on the same finite-dimensional vector space) and if , then is not nilpotent.

Exercise 3. Prove that if is a nilpotent linear transformation of index on a finite-dimensional vector space, then for .

Exercise 4. If is a linear transformation (on a finite-dimensional vector space over an algebraically closed field), then there exist linear transformations and such that , is diagonable, is nilpotent, and ; the transformations and are uniquely determined by these conditions.

EXERCISES

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