5.III. Nilpotence

5.III.1 Self-Composition

5.III.2 Strings

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

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/dxFor the derivative map d/dx: 3 → 3 given by

/2 3 22 3

d dx

a bx cx dx b c dx

the second power is the second derivative:2 2/

2 3 2 6d dx

a bx cx dx c dx

the third power is the third derivative:3 3/

2 3 6d dx

a bx cx dx d d n/ dxn = zero map n > 3.

Example 1.2 :

0

ta b b a

c d d

2

0 0

ta b a b

c d

3

0 0

ta b b a

c d

2 2nt t 2 1 3nt t 1n

Lemma 1.3:

For any transformation t : V → V ,

the rangespaces of the powers form a descending chain

V ( t ) ( t2 ) …

and the nullspaces form an ascending chain

{0} ( t ) ( t2 ) …

Further, k s.t. for powers less than k the subsets are proper

( if j < k then ( t j ) ( t j+1 ) and ( t j ) ( t j+1 ) ),

while for powers greater than k the sets are equal

( if j k then ( t j ) = ( t j+1 ) and ( t j ) = ( t j+1 ) ).

Proof :

For any map, dim(rangeSpace) + dim(nullSpace) = dim(domain).

→ rangeSpace & nullSpace must grow at the expense of each other.

Let w( t j+1 ), i.e., w = t j+1(v),

then w = t j ( t(v) ) so that w( t j ).

( t j ) ( t j+1 )

The nullSpace portion can be similarly proved.

Next, suppose ( t k ) = ( t k+1 ), then

t : ( t k+1 ) → ( t k+2 )

becomes the map t with domain ( t k ) so that its range

( t k+2 ) = ( t k+1 ) = ( t k )

This proves the “further” part of the lemma.

Example 1.4:The derivative map

/2 3 22 3

d dx

a bx cx dx b c dx

has this chain of rangespaces: 3 2 1 0 0 0P P P P

and this chain of nullspaces: 0 1 2 3 3 3 0 P P P P P P

Example 1.5:The projection π: 3 → 3 by

1 1

2 2

3 0

c c

c c

c

has this chain of rangespaces: 3 2 R R C

and this chain of nullspaces: 2 0 N N

Example 1.6:Let t : 2 → 2 be the map

On iteration the rangespace shrinks:

20 1 2 0 22c c x c x c c x

02t R P 1t R P 2

0t R P 3t R

while the nullspace grows

0t 0N t cx c N C 21t N P 3t N

Definition 1.7:

Let t be a transformation on an n-D space.

The generalized rangespace (or the closure of the rangespace) is (t) = (t n ).

The generalized nullspace (or the closure of the nullspace) is (t) = (t n ).

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) 2 2:s P P 2 2a bx cx b cx by

(b) 2 2:t P P 2 2a bx cx b cx ax by

5.III.2. Strings

Lemma 2.1:

Let t : V → V be a linear transformation, then

V = (t) (t)

i.e., dim(V ) = dim( (t) ) + dim( (t) )

and (t) (t) = { 0 }.

Proof:

The statement dim(V ) = dim( (s) ) + dim( (s) )holds for any transformation s : V → V. Hence, it holds for s = t k k.

Let v (t) (t) = (t n ) (t n ), where (t n ) = (t n+1 ).

→ t n ( v ) = 0 since v (t n )

t : (t) → (t) is 1-1 since (t n ) = (t n+1 ).

t n ( v ) = 0 → v = 0. QED

t : V → V and t : (t) → (t) are different maps if V (t).

t : (t) → (t) is called the restriction of t to (t).

This restriction map is 1-1 onto and hence nonsingular.

Note: V = V { 0 } = ( t 0 ) ( t 0 )

Example 2.3:

2 2:n C C by1 0

0 1

and0 0

1 0

2 2

0 0

1 0n

NE E

Since e2 is in both (n) and (n), 2 (n) (n).

The action of n can be depicted by a string 1 2e e 0

Example 2.4: 1-string transformation

4 4:n 4C C be defined by the stringLet1 2 3 4 e e e e 0

→4 4

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

n

NE E

Thus (n) (n) = span { e4 }

(n2) (n2) = span { e3 , e4 }

(n3) (n3) = span { e4 }

(n4) (n4) = { 0 }

2

0 0 0 0

0 0 0 0

1 0 0 0

0 1 0 0

N 3

0 0 0 0

0 0 0 0

0 0 0 0

1 0 0 0

N

Example 2.5: 2-string transformation

Let t be defined wrt = β1 , … , β5 by 2 strings

1 2 3β β β 0 4 5β β 0

→

0 0 0 0 0

1 0 0 0 0

0 1 0 0 0

0 0 0 0 0

0 0 0 1 0

t

TB B

Definition 2.6:A nilpotent transformation is one with a power that is the zero map. A nilpotent matrix is 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.

Example 2.8:

The differentiation map d/dx: 2 → 2 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 = x2, 2x, 2 gives this representation:

0 0 0

1 0 0

0 1 0

d

dx

TB B

Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones.

Example 2.9: (see Example 2.4)

4 4:n 4C C defined by the string

1 2 3 4 e e e e 0 → 4 4

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

n

NE E

Let

1 0 1 0

0 2 1 0, , ,

1 1 1 0

0 0 0 1

D

→ 1n M P NPD D

4

1 0 1 0

0 2 1 0

1 1 1 0

0 0 0 1

id

PD E

1 4 1 0

0 2 1 0

1 4 1 0

1 1 1 0

Index of nilpotency is invariant under similarity transform since

1 1n n P NP P N P

Definition 2.10: t-string basisLet t be a nilpotent transformation on V . A t-string generated by v V is a sequence v , t(v), …, t k1 (v) .This sequence has length k.A t-string basis is a basis that is a concatenation of t-strings.

Example 2.11:In Example 2.5, the t-strings 1 2 3β β β 0 4 5β β 0

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.

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.

Next, we shall show that every nilpotent map has an associated string basis.

Possible counter-example:

t : 5 → 5 by5 5

0 0 0 0 0

0 0 0 0 0

1 1 0 0 0

0 0 0 0 0

0 0 0 1 0

t

TE E

Let the string basis, if exists, be = β1 , … , β5 .

Since the index of nilpotency is 2, at least one string in the basis has length 2.

There’re only 2 possibilities:

1 2

3 4

5

β β 0

β β 0

β 0

1 2

3 4 5, ,

β β 0

β β β 0

or

Nullity = 3 Nullity = 4

0 0 0 0 0

0 0 0 0 0

1 1 0 0 0

0 0 0 0 0

0 0 0 1 0

TThe nullspace of is , ,

0

x

x

t x z rz

r

N

xz

x

C

Nullity of t is 3 so that a t-string basis must be of form1 2

3 4

5

β β 0

β β 0

β 0

First, β2 & β4 are picked from (t) (t), e.g.,

2

0

0

1

0

0

β 4

0

0

0

0

1

β

β5 is picked from what’s left in (t), e.g., 5

1

1

0

0

0

β

Finally, β1 & β3 are chosen s.t. t(β1 ) = β2 and t(β3 ) = β4 , i.e.,

1

1

0

0

0

0

β 3

0

0

0

1

0

β

→

0 0 0 0 0

1 0 0 0 0

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

is of the canonical form

5

1 0 0 0 1

0 0 0 0 1

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

id

PB E

1t P TPB B

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.

Example 2.15:

1 1

1 1

M

→ Index of nilpotency of M = 2.

2 2MN C2 0 0

0 0

M

xx

x

MN C

t-string of M: β1 β2 0.0 0

1 0m

B B

In terms of 2 : 2

1

1

β 1

1

0

β

2

1 1

0 1id

PB E

1m P MPB B

1 1 1 1 1 1

0 1 1 1 0 1

0 0

1 0

Example 2.16:

0 0 0 0 0

1 0 0 0 0

1 1 1 1 1

0 1 0 0 0

1 0 1 1 1

N

0

0

,u vu v

u

v

NN C

2

0 0 0 0 0

0 0 0 0 0

1 0 0 0 0

1 0 0 0 0

0 0 0 0 0

N 2

0

, , ,

y

y z u vz

u

v

NN C

3 N O 3 5NN C

t-strings: 1 2 3

4 5

β β β 0

β β 0

First, β3 & β5 are picked from (n) (n), e.g., 3

0

0

1

1

0

β 5

0

0

1

0

1

β

β2 & β4 are are chosen s.t. n(β2 ) = β3 and n(β4 ) = β5 , i.e., 2

0

1

0

0

0

β4

0

0

1

0

0

β

β1 is chosen s.t. n(β1 ) = β2 , i.e., 1

1

0

1

0

0

β

Hence, the block subdiagonal form is

5

1 0 0 0 0

0 1 0 0 0

1 0 1 1 1

0 0 1 0 0

0 0 0 0 1

id

PB E

1

0 0 0 0 0

1 0 0 0 0

0 1 0 0 0

0 0 0 0 0

0 0 0 1 0

n

P NPB B

1 2 3

4 5

β β β 0

β β 0

where

Exercises 5.III.2.

1. Find the canonical form of these matrices.

1 1

2 21 1

2 2

1 1 1

1 0 1

1 1 1

2. Show that the only eigenvalue of a nilpotent matrix is zero.