5.III. Nilpotence

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##### 5.III. Nilpotence

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1. 5.III. Nilpotence Every square matrix is similar to one that is a sum of 2 kinds of simple matrices: diagonal & nilpotent matrices. 5.III.1 Self-Composition 5.III.2 Strings

2. 5.III.1. Self-Composition Since its domain & codomain are the same, a linear transformation t : V → V can be iterated, i.e., self-compositions such as t2 = t t and t3 = t t  t are defined. Example 1.1: d/dx For the derivative map d/dx: P3 →P3 given by the second power is the second derivative: the third power is the third derivative: d n/ dxn = zero map  n > 3.

3. Example 1.2 : Lemma 1.3: For any transformation t : V → V , the rangespaces of the powers form a descending chain V R( t )  R( t2 )  … and the nullspaces form an ascending chain {0}  N( t )  N( t2)  … Further,  k s.t. for powers less than k the subsets are proper ( if j < k then R( t j) R( t j+1 ) and N( t j) N( t j+1 ) ), while for powers greater than k the sets are equal ( if j  k then R( t j) =R( t j+1 ) and N( t j) =N( t j+1 ) ).

4. Proof : For any map, dim(rangeSpace) + dim(nullSpace) = dim(domain). → rangeSpace & nullSpace must grow at the expense of each other. Let wR( t j+1 ), i.e., w = t j+1(v), then w = t j ( t(v) ) so that wR( t j).  R( t j)  R( t j+1 ) The nullSpace portion can be similarly proved. Next, suppose R( t k) = R( t k+1 ), then t : R( t k+1 ) → R( t k+2 ) becomes the map t with domain R( t k) so that its range R( t k+2 ) = R( t k+1 ) = R( t k) This proves the “further” part of the lemma.

5. Example 1.4: The derivative map has this chain of rangespaces: and this chain of nullspaces: Example 1.5: The projection π: C3 →C3 by has this chain of rangespaces: and this chain of nullspaces:

6. Example 1.6: Let t : P2 →P2 be the map On iteration the rangespace shrinks: while the nullspace grows

7. Definition 1.7: Let t be a transformation on an n-D space. The generalized rangespace(or the closure of the rangespace) is R(t) = R(t n). The generalized nullspace(or the closure of the nullspace) is N(t) = N(t n).

8. Exercises 5.III.1. 1. For each map, give the chain of rangespaces and the chain of nullspaces, and the generalized rangespace and the generalized nullspace. (a) by (b) by

9. 5.III.2. Strings Lemma 2.1: Let t : V → V be a linear transformation, then V = R(t) N(t) i.e., dim(V ) = dim( R(t) ) + dim( N(t) ) and R(t)  N(t) = { 0 }. Proof: The statement dim(V ) = dim( R(s) ) + dim( N(s) ) holds for any transformation s : V → V. Hence, it holds for s = t k k. Let v R(t)  N(t) = R(t n)  N(t n), where R(t n) = R(t n+1). → t n ( v ) = 0 since v N(t n) t : R(t) → R(t) is 1-1 since R(t n) = R(t n+1).  t n ( v ) = 0 → v = 0. QED

10. t : V → V and t : R(t) → R(t) are different maps if V R(t). t : R(t) → R(t) is called the restriction of t to R(t). This restriction map is 1-1 onto and hence nonsingular. Note: V = V  { 0 } = R( t 0) N( t 0 ) Example 2.3: by and  Since e2 is in both R(n) and N(n), C2 R(n) N(n). The action of n can be depicted by a string

11. Example 2.4: 1-string transformation Let be defined by the string → Thus R(n) N(n) = span { e4 } R(n2) N(n2) = span { e3 , e4 } R(n3) N(n3) = span { e4 } R(n4) N(n4) = { 0}

12. Example 2.5: 2-string transformation Let t be defined wrt B =  β1 , … , β5  by 2 strings → Definition 2.6: A nilpotenttransformation is one with a power that is the zero map. A nilpotent matrixis one with a power that is the zero matrix. In either case, the least such power is the index of nilpotency. Example 2.7: The index of nilpotency in Examples 2.3, 2.4, 2.5 are 2, 4, 3, resp.

13. Example 2.8: The differentiation map d/dx: P2 →P2 is nilpotent of index 3 ( the 3rd derivative of any quadratic polynomial is zero. ) This map’s action is described by the string x2 2x  2  0 and taking the basis B=  x2, 2x, 2 gives this representation:

14. Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones. Example 2.9: (see Example 2.4) defined by the string → Let → Index of nilpotency is invariant under similarity transform since

15. The rest of this subsection aims to prove Theorem 2.13, which shows that every nilpotent matrix is similar to one that is all zeros except for blocks of subdiagonal ones. Definition 2.10: t-string basis Let t be a nilpotent transformation on V . A t-string generated byv V is a sequence  v , t(v), …, t k1(v) . This sequence has length k. A t-string basisis a basis that is a concatenation of t-strings. Example 2.11: In Example 2.5, the t-strings & are of length 3 & 2, and can be concatenated to make a basis for the domain of t. Lemma 2.12: If a space has a t-string basis then the longest string in it has length equal to the index of nilpotency k of t. Proof: t k (v) = 0 → longest t-string has length  k. If all t-strings have length < k, then  v s.t. t k(v) 0. QED.

16. Next, we shall show that every nilpotent map has an associated string basis. Possible counter-example: t : C5 → C5 by Let the string basis, if exists, be B =  β1 , … , β5 . Since the index of nilpotency is 2, at least one string in the basis has length 2. There’re only 2 possibilities: or Nullity = 4 Nullity = 3

17. The nullspace of is  Nullity of t is 3 so that a t-string basis must be of form First, β2 & β4 are picked from R(t)  N(t), e.g.,

18. β5 is picked from what’s left in N(t), e.g., Finally, β1 & β3 are chosen s.t. t(β1 ) = β2 and t(β3 ) = β4 , i.e., → is of the canonical form

19. Theorem 2.13: Any nilpotent transformation t is associated with a t-string basis. While the basis is not unique, the number and the length of the strings is. Proof, by induction: Skipped: see Hefferon, p.375. Corollary 2.14: Every nilpotent matrix is similar to a matrix that is all zeros except for blocks of subdiagonal ones (canonical form, if properly ordered, of the similarity class). That is, every nilpotent map is represented wrt some basis by such a matrix.

20. Example 2.15: → Index of nilpotency of M = 2. t-string of M: β1 β2 0. In terms of E2 :

21. Example 2.16:

22. t-strings: First, β3 & β5 are picked from R(n)  N(n), e.g., β2 & β4 are are chosen s.t. n(β2 ) = β3 and n(β4 ) = β5 , i.e., β1 is chosen s.t. n(β1 ) = β2 , i.e.,

23. Exercises 5.III.2. 1. Find the canonical form of these matrices. 2. Show that the only eigenvalue of a nilpotent matrix is zero.