On commutativity and lie nilpoten y in matrix algebras

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by

Mahlare Gerald Sehoana

Thesis presented in partial fullment of the requirements for

the degree of Master of S ien e (Mathemati s) in the Fa ulty

of S ien e at Stellenbos h University

Supervisor: Prof. L. vanWyk

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De laration

By submittingthis thesis ele troni ally,I de lare that the entirety of the work ontained

therein is my own, original work, that I am the sole author thereof (save to the extent

expli itly otherwise stated), that reprodu tion and publi ation thereof by Stellenbos h

University will not infringe any third party rights and that I have not previously in its

entirety orin part submitted itfor obtainingany quali ation.

2015/09/30

Date: ...

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Abstra t

On Commutativity and Lie nilpoten y in Matrix Algebras

M.G.Sehoana

Department of Mathemati al S ien es,

Stellenbos h University,

Private Bag X1, Matieland 7602, South Afri a.

Thesis: MS (Mathemati s)

September 2015

In thisthesis we rstdis uss the proofbyMirzakhani [9℄ofS hur's Theorem whi hgives

the maximumnumberoflinearlyindependentmatri esina ommutativealgebraof

n × n

matri es over a eld

F

. An example illustrating the appli ation of S hur's Theorem is given.

Se ondly,wedis ussthe Cayley-HamiltonTheoremwhi hassertsthatany

n×n

matrix

A

satisesits hara teristi polynomial. Adedu tionofaCayley-Hamiltontra eidentityfor

a

2 × 2

matrix

A

over a ommutativering fromthe Cayley-HamiltonTheorem is shown. Wethen dis uss the Cayley-Hamilton tra e identity forany matrix

A ∈ M

2 (R)

when (i)

R

is ommutative,

(ii)

R

isnot ne essarily ommutative,

(iii)

R

is not ne essarily ommutative and

tr(A) = 0

,

(iv)

R

is not ne essarily ommutativeand satises the identity

[[x, y], [x, z]] = 0.

Lastly,wedis ussthematrixalgebras

U

∗

n

(R)

,inparti ularthematrixalgebras

U

∗

3 (R)

and

U

∗

4 (R)

, in relation to polynomial identities

[[. . . [[x

1 , x

2 ], x

3 ], . . .], x

n

] = 0

,

[x, y][w, z] = 0

and

[[x, y], [w, z]] = 0

.

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Uittreksel

Oor Kommutatiwiteit en Lie nilpotensie in Matriksalgebras

(On Commutativityand Lienilpoten y inMatrix Algebras)

M.G.Sehoana

Departement Wiskundige Wetenskappe,

UniversiteitStellenbos h,

PrivaatsakX1, Matieland 7602, Suid Afrika.

Tesis: MS (Wiskunde)

September 2015

In hierdie tesis beskryf ons eerstens die bewys deur Mirzakhani [9℄ van S hur se Stelling

wat die maksimum aantal lineêr onafhanklike matrikse in 'n kommutatiewe algebra van

n × n

matrikse oor 'n liggaam

F

gee. 'n Voorbeeld word gegee wat die toepassing van S hur seStelling illustreer.

TweedensbespreekonsdieCayley-HamiltonStellingwatbeweer datelke

n × n

matriks

A

sykarakteristiekepolinoombevredig. 'nAeidingvan'nCayley-Hamiltonspooridentiteit

vir 'n

2 × 2

matriks

A

oor 'n kommutatiewe ring vanuit die Cayley-Hamilton Stelling word gegee. Ons bespreek dan die Cayley-Hamilton spoor identiteit vir enige matriks

A ∈ M

2 (R)

wanneer (i)

R

kommutatief is,

(ii)

R

nienoodwendig kommutatief is nie,

(iii)

R

nie noodwendig kommutatief is nieen sp

(A) = 0

,

(iv)

R

nienoodwendigkommutatief isnie en dieidentiteit

[[x, y], [x, z]] = 0

bevredig.

Laastensbespreekons diematriksalgebras

U

∗

n

(R)

,inbesonder diematriksalgebras

U

∗

3 (R)

en

U

∗

4 (R)

, met betrekking totdiepolinoomidentiteite

[[. . . [[x

1 , x

2 ], x

3 ], . . .], x

n

] = 0

,

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A knowledgements

I would like toexpress my sin ere gratitude to the following people and organisations:

⋄

Prof LeonvanWyk forthe guidan e, dire tionand enlightenmentthroughout my stud-ies. His patien e and onstru tive omments kept my hopes alive. He brought ba k my

love forMathemati s.

⋄

Prof David Kubayi for getting me started on this proje t and for the support that, together with the Deanof FSA at UL, provided duringhis time asHOD.

⋄

My family for theirmoral support and motivation.

⋄

Department of Mathemati al S ien es at Stellenbos h University for allowing me the opportunity tostudy for anMS degree.

⋄

Fellow students Sandile Mkhaliphi and Andry Rabenantoandro for assisting me with Latex.

⋄

Colleagues in the s hool of Mathemati al and Computer S ien es at the University of Limpopofor their en ouragement.

⋄

DST-NRF Centre of Ex ellen e in Mathemati al and Statisti alS ien es (CoE-Mass) for their nan ialsupport.

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Dedi ations

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Contents De laration i Abstra t ii Uittreksel iii A knowledgements iv Dedi ations v Contents vi 1 Introdu tion 1

2 Mirzakhani's Simple Proof of S hur's Theorem 5

3 Cayley-Hamilton Theorem 14

4 Commutativity and Lie nilpoten y in the Matrix Algebra

U

∗

n

(R)

21 4.1 The ring

U

∗

3 (R)

. . . 21 4.2 The ring

U

∗

4 (R)

. . . 34

5 Cayley-Hamilton Tra e Identity for

2 × 2

Matri es 45 5.1 Tra e identities for

2 × 2

matri es . . . 45 5.2 Relationshipbetween identities . . . 55

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Chapter 1

Introdu tion

This hapter is mainlyabrief ba kgroundand overview of the subsequent Chapters

2

to

5

. Also inthis hapterwedis uss some of the on epts whi h we have used anumber of times in this thesis. Where possible, we have supplied examples to substantiate laims

made. Chapter

5

anbeviewed asa ontinuationof Chapter

3

. Though inChapter

3

we dealwithmatri esovertheeld

R

ofrealnumbersandinChapter

5

wedealwithmatri es over an arbitrary ring

R

. Coin identally, Chapters

2

and

4

deal with upper triangular matri es. InChapter

2

ween ounter uppertriangularmatri eswhi hmutually ommute whereas in Chapter

4

ommutativity of the said matri es isnot ne essary.

A binary operation

∗

ona set

A

issaid to be ommutativeif and onlyif

x ∗ y = y ∗ x

for all

x, y ∈ A.

Were all that aring

R

isan algebrai stru ture with two binary operations alled additionand multipli ation. One of the axioms of a ring

R

states that addition is ommutative. However, ina ring

R

multipli ationis not ne essarily ommutative. Thus, if multipli ation is ommutative su h a ring is alled a ommutative ring. A nonempty

subset

B

of a ring

R

is said tobe a subringof

R

if

B

is itself a ring with respe t to the operations of addition and multipli ation in

R

. Fora nonempty subset

B

of a ring

R

to be a subring it is su ient that

ab ∈ B

and

a − b ∈ B

for all

a, b ∈ B.

The olle tion

M

n

(R)

,where

R

is aring, of all

n × n

matri eshaving elements of

R

asentries is aring. For

n ≥ 2

,

M

n

(R)

isnot a ommutativering. But, for a ommutativering

R

,the ring

B =

a b

0 a

| a, b ∈ R

is ommutativeand

B

is asubringof

M

2 (R)

. Foreveryring

R

the trivialsubringB={0} is ommutative. Thus every ring has at least one subring whi h is ommutative.

In Chapter

2

we shall en ounter a maximal ommutative subalgebraof

M

n

(F )

whi hwe use to illustrate anappli ation of S hur's Theorem. (A subring

B

of a ring

R

is said to bea maximalsubringwithrespe t toproperty

Y

if

B 6= R

and thereexists nosubring

C

in

R

with the property

Y

su h that

B ⊂ C ⊂ R

.) In fa t, a ording toS hur's Theorem the numberoflinearly independentmatri esinthe subalgebra

B

above is

⌊2

2 _{/4⌋ + 1 = 2}

.

Moreover, we see that the number of elements in a basis for

B

is

2

. A basis of a ve tor spa e

V

is a subset

W ⊆ V

whi h is linearly independent and spans

V

. A basis of a

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CHAPTER 1. Introdu tion 2

ve tor spa e isa maximal linearlyindependent subset of that ve tor spa e.

The expression

⌊x⌋

represents the greatest integer less than or equal to

x

. If

x

is an integer, we have

⌊x⌋ = x

and

⌊x + b⌋ = x

for

0 < b < 1.

The hara teristi polynomialof an

n × n

matrix

A

overa ommutativering

R

isdened tobe

p(λ) =

det

(A − λI) = k

n

λ

n

_{+ k}

n−1

λ

n−1

+ . . . + k

1 λ + k

0 ,

k

i

∈ R.

Bythe tra eofa squarematrix

A

,denoted

tr(A)

, ismeantthe sum ofthe entries onthemaindiagonalof

A

,fromtheupperlefttothelowerright. Thefollowingproperties of tra es of any matri es

A, B ∈ M

n

(R)

are used in this thesis:

(i)

tr(A ± B) = tr(A) ± tr(B),

(ii)

tr(cA) = c tr(A),

where

c

isa onstant.

Thenontrivial oe ientsofthe hara teristi polynomialofamatrix

A

an beexpressed expli itly in terms of tra es of powers of A, (see [15℄, [13℄). In [13℄, it is shown that if

n = 4

, then

p(λ) = λ

4 − T

1 λ

3 +

1

2 T

2

1 − T

2 λ

2 −

1

6 T

3

1 − 3T

1 T

2 + 2T

3 λ

+

1

24 T

4

1 − 6T

1

2 T

2 + 8T

1 T

3 + 3T

2

2 − 6T

4 = λ

4 − tr(A)λ

3 +

1

2 tr

2 _{(A) − tr(A}

2 _)λ

2 −

1

6 tr

3 _{(A) − 3 tr(A) tr(A}

2 _{) + 2 tr(A}

3 _)λ

+

1

24 tr

4 _{(A) − 6 tr}

2 _{(A) tr(A}

2 _{) + 8 tr(A) tr(A}

3 _{) + 3 tr}

2 _(A

2 _{) − 6 tr(A}

4 _),

where

T

m

= tr(A

m

_).

In this thesis we deal with the ase when

n = 2

, that is,

p(λ) = λ

2 − tr(A)λ +

1

2 tr

2 _{(A) − tr(A}

2 ₎

and obviously repla ing

λ

by

A

and introdu ingthe identity matrix

I ∈ M

2 (R)

yields

p(A) = A

2 − tr(A)A +

1

2 tr

2 _{(A) − tr(A}

2 _)I.

The equation just given leads to the Cayley-Hamilton tra e identity for a

2 × 2

matrix whi hwe dis uss inChapter

5

under various hypotheses with

1

2 ∈ R.

Denition 1. Let

R

be aring (not ne essarily ommutative) and

a, b ∈ R

. An element of the form

[a, b] = ab − ba

is alled a ommutator, more pre isely, the ommutator of

a

and

b.

Denition2. Aring

R

is alledLienilpotentofindex

n (n ≥ 2)

if

R

satisestheidentity

[[[. . . [[x

1 , x

2 ], x

3 ], . . .], x

n

], x

n+1

] = 0

but not the identity

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Denition 3. A ring

R

is alled Lie nilpotent if it is Lie nilpotent of index

n

for some

n ≥ 2.

(Wenotethata ommutativeringmaybe alleda"Lienilpotentringofindex 1".)

The on eptsof ommutativityandLienilpoten yare"somehow"relatedinthesensethat

one implies the other. Commutativity always implies Lie nilpoten y and Lie nilpoten y

implies ommutativity only if the on erned ring is of Lie nilpotent index 1. Thus, we

an say ommutativity is a stronger ondition than Lie nilpoten y. For example, in a

ommutative ring the identity

[x

1 , x

2 ] = 0

holds whi h in turn implies

[[x

1 , x

2 ], x

3 ] = 0

, whi hinturn implies

[[[x

1 , x

2 ], x

3 ], x

4 ] = 0

andso on. Weshall see inProposition33that the ring

U

∗

3 (R)

satises the hypothesis for Lie nilpoten y but by Example 30,

U

∗

3 (R)

is not ommutative.

Wefurthernoti ethat inthe algebras

U

∗

n

(R)

wedis uss inthisthesis,

R

isrequiredtobe ommutative in order to attain Lie nilpoten y. This fa t is observed in Theorem 44. It

is shown in Example 45that Lienilpoten y may not ne essarily beattained in

U

∗

n

(R)

if

R

is non ommutative. (In ontrast, a subring

V

∗∗

n

(R) ⊂ U

n

∗

(R)

, with a non ommutative ring

R

,dened in Chapter

4

isLie nilpotentof index

2

,for all

n

. The subring

V

∗∗

n

(R)

is a tually a version of the subring

F

n

given in Chapter

2

but with a non ommutative ring

R

.)

We see again the fundamental role of ommutativity in Corollary 37 where an algebra

U

∗

3 (U

3 ∗

(R))

, with non ommutative ring

U

∗

3 (R)

and ommutative

R

, satises the identity

[[x, y], [w, z]] = 0

and does not satisfy either of the stronger identities

[[x, y], z] = 0

and

[x, y][w, z] = 0

. On the other hand,the algebra

U

∗

4 (U

3 ∗

(R))

with ommutative

R

doesnot satisfy

[[x, y], [w, z]] = 0

but

[[[x, y], [w, z]], [[u, v], [r, s]]] = 0

.

Fora ommutativering

R

,anyprodu t

[x

1 , y

1 ][x

2 , y

2 ]

in

U

∗

n

(R)

,

n ≤ 4,

isequaltozeroand the dis ussion just after Remark 48 shows that for

n ≥ 7

a produ t

[x

1 , y

1 ][x

2 , y

2 ][x

3 , y

3 ]

may not ne essarilybe equaltozero. This leadsto Theorems52 (ii)and 53whi h give a

smallest value of

k

forwhi h any produ t of the form

[x

1 , y

1 ][x

2 , y

2 ] · · · [x

k

, y

k

]

(1.1) in

U

∗

n

(R), R

ommutative,isequal tozero. This value of

k

dependson

n

. When

R

isnot ommutative, (1.1) may not ne essarilybe equalto zero for su h values of

k

as shown in Remark 56(i.e, thereare matri esin

U

∗

n

(R)

for whi h

[x

1 , y

1 ][x

2 , y

2 ] · · · [x

k

, y

k

] 6= 0

). Denition4. Let

R

beanarbitraryring. Thematrixunit

E

i,j

in

M

n

(R)

with

1 ≤ i, j ≤ n,

is dened to bethe matrix with

1

in position

(i, j)

and zeroselsewhere.

Notethat

E

i,j

E

k,l

=

E

i,l

if

j = k,

0

if

j 6= k.

Alternatively

E

i,j

E

k,l

= δ

j,k

E

i,l

,

where

δ

j,k

=

1

if

j = k,

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CHAPTER 1. Introdu tion 4

The fun tion

δ

j,k

is alled the Krone ker delta. We observe that for

a, b ∈ R,

(aE

i,j

)(bE

k,l

) = (ab)E

i,j

E

k,l

=

(ab)E

i,l

if

j = k,

0

if

j 6= k.

(1.2)

It should be noted that in all the examples that show that ertain statements do not

ne essarilyhold when

R

is non ommutative, we used spe i matri esin the ring

U

∗

n

(R)

be ause there may be matri esin

U

∗

n

(R)

for whi h the statements hold. For example, if

R

is non ommutative,a produ tof the form(1.1) an be zero as in

[E

1,2

, E

1,3

][A

2 , B

2 ] · · · [A

k

, B

k

] = 0,

for all

k

and

E

1,2

, E

1,3

, A

k

, B

k

in

U

∗

n

(R)

. Wehave

[E

1,2

, E

1,3

] = 0

,a ording toDenition 1 and the observation that follows afterDenition 4.

Lastly, atthe end of Chapter 5 we dis uss the proof of the fa tthat in a ring ontaining

1

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Chapter 2

Mirzakhani's Simple Proof of S hur's

Theorem

Foraeld

F

, the ring

M

n

(F )

of

n × n

matri eswith

n > 1

andelementsin

F

is non om-mutative. But there are subalgebras

S ⊂ M

n

(F )

whi h are ommutative. Theorem

10

in this hapterknown astheTheoremofS hurdealswithsubalgebrasof

M

n

(F )

ofmutually ommutative matri es. It gives the maximum number of linearly independent matri es

in a maximal ommutative subalgebra. Mirzakhani [9℄ provided a simpler proof, using

indu tion, of this theorem whi hwe disse t inthis hapter. Mirzakhani [9℄alsoprovided

anexample of a ommutative subalgebrasatisfying the hypothesis of Theorem10.

Denition 5. A set

X = {x

1 , x

2 , . . . , x

n

}

is said to be a spanning set of a ve tor spa e

W

overa eld

F

if every

w ∈ W

is alinear ombinationof the

x

i

∈ X

, i.e.,

w = a

1 x

1 + a

2 x

2 + · · · + a

n

x

n

where

a

i

∈ F

.

Denition 6. A set

V = {v

1 , v

2 , . . . , v

r

}

of non-zero ve tors in a ve tor spa e over eld

F

issaid to be linearly independent whenever

a

1 v

1 + a

2 v

2 + · · · + a

r

v

r

= 0

implies that

a

i

= 0

for all

i

, where

a

i

∈ F.

Denition 7. Let

A = [A

1 |A

2 | . . . |A

n

]

be an

m × n

matrix over a eld

F

, with the

A

i

's as olumns of

A

. The rank of

A

is dened to be the maximum number of linearly independent ve tors inthe set

{A

1 , A

2 , . . . , A

n

}.

Denition 8. Let

A

be an

m × n

matrix over a eld

F

. The null spa e of

A

is the set ofall

x

in

F

n

su hthat

Ax = 0.

Thedimension ofthe nullspa e of

A

is alledthe nullity of

A

.

Remark 9. The nullity of a matrix

A

is the dimension of the solution spa e of

Ax = 0

, whi h is the same as the number of parameters in the general solution of

Ax = 0

and whi his the same as the number of freevariables.

(13)

CHAPTER 2. Mirzakhani'sSimple Proof of S hur's Theorem 6

Theorem 10. Let

A

be an

m × n

matrix. Then

rank A + nullity A = n.

AsubmatrixofanygivenmatrixAisamatrixwhi hisobtained fromAby removingany

numberofrows and/or olumns. Amatrixissaidtobepartitionedwheneveritisdivided

into submatri esby drawing verti al and horizontallines between its rows and olumns.

Example 11. If

A =







a

1,1

a

1,2

· · · a

1,n

a

2,1

a

2,2

· · · a

2,n

. . . . . . . . . . . .

a

m,1

a

m,2

· · · a

m,n







,

thenapartitioningof

A

intosubmatri es

B, C, D

and

E

oforder

3×2, 3×(n−2), (m−3)×2

and

(m − 3) × (n − 2)

respe tively is







a

1,1

a

1,2

a

1,3

· · · a

1,n

a

2,1

a

2,2

a

2,3

· · · a

2,n

a

3,1

a

3,2

a

3,3

· · · a

3,n

a

4,1

a

4,2

a

4,2

· · · a

4,n

. . . . . . . . . . . . . . .

a

m,1

a

m,2

a

m,3

· · · a

m,n







=

B C

D E

where

B, C, D

and

E

an bethought of aselements of

A

. We note that the partitioning of a matrix intosubmatri es is not unique.

Denition 12. A eld

F

is said tobealgebrai ally losed if every polynomialequation

a

0 + a

1 x + a

2 x

2 + · · · + a

n

x

n

= 0

with oe ients in

F

has a solutionin

F

.

Theeld

R

ofrealnumbersisnot algebrai ally losedastheequation

x

2 _{+ 1 = 0}

doesnot

havea solutionin

R

.

Butthe eld

C

of omplexnumbers isalgebrai ally losed, (see[6 ℄).

Westate, withoutproof,the followingtheorem (whi h an befound in[4℄) be auseof its

signi an e inthe proof of Theorem 14.

Theorem 13. Let

F

n

be a family of ommuting matri es of order

n

over analgebrai ally losed eld

F

. Then there exists a nonsingular matrix

P

of order

n

with entries in

F

su h that

P

−1

_F

(14)

Theorem 14. The maximum number of mutually ommuting linearly independent

ma-tri es of order

n

over a eld

F

is

⌊n

2 _{/4⌋ + 1}

.

Proof. The proof is by mathemati alindu tion.

(i)For

n = 1

the statementis obviously true.

(ii) Assume the theorem is true for

n − 1,

i.e., a set onsisting of mutually ommuting matri es of order

(n − 1) × (n − 1)

has at most

⌊(n − 1)

2 _{/4⌋ + 1}

linearly independent

matri es.

(iii)Let

F

n

bea familyof ommutingmatri esof order

n

overa eld

F

. Suppose

F

n

has more than

⌊n

2 _{/4⌋ + 1}

linearly independent matri es. Assume, withoutloss of generality,

that

F

is algebrai ally losed, then by Theorem 9 there exists a nonsingular matrix

P

withentries in

F

su hthat

P

−1

_F

n

P

isafamilyofuppertriangularmatri es. Alsoforany

T, U ∈ F

n

,

(P

−1

UP )(P

−1

T P ) = P

−1

U(P P

−1

)T P = P

−1

(UT )P

and

(P

−1

T P )(P

−1

UP ) = P

−1

T (P P

−1

)UP = P

−1

(T U)P.

But

UT = T U

and thus

(P

−1

_{UP )(P}

−1

_{T P ) = (P}

−1

_{T P )(P}

−1

_{UP )}

showing that the

ma-tri esin

P

−1

_F

n

P

aremutually ommuting. Let

V

bethe ve torspa espanned by the set

P

−1

_F

n

P

. Then

V

onsists of uppertriangularmatri eswhi h ommute. Thus,

P

−1

_F

n

P ⊆

span

{P

−1

_F

n

P } = V

and dim

V ≥ ⌊n

2 _{/4⌋ + 2.}

Sin e a subset of a linearly independent set is also linearly

independent there exists a linearly independent subset

{A

1 , A

2 , . . . , A

q

}

of a basis of

V

,

q = ⌊n

2 _{/4⌋ + 2}

. Sin e ea h of the

A

i

's is an uppertriangularmatrix, the partitioning of ea h

A

i

intosubmatri es

M

i

,

H

i

and a zero matrix of order

(n − 1) × (n − 1)

,

1 × n

and

(n − 1) × 1

respe tively,yields matri esof the form

A

i

=







H

i

0

0 M

i

. . .

0 





.

Now, partitioningthe

H

i

further (where

L

i

, N

i

and

M

i

aresubmatri es oforder

1 × 1, 1 ×

(n − 1)

and

(n − 1) × (n − 1)

respe tively) we have

A

i

A

j

=

L

i

N

i

0 M

i

L

j

N

j

0 M

j

=

L

i

L

j

L

i

N

j

+ N

i

M

j

0 M

i

M

j

and

A

j

A

i

=

L

j

N

j

0 M

j

L

i

N

i

0 M

i

=

L

j

L

i

L

j

N

i

+ N

j

M

i

0 M

j

M

i

.

(15)

CHAPTER 2. Mirzakhani'sSimple Proof of S hur's Theorem 8

Sin e

A

i

A

j

= A

j

A

i

it follows that

M

i

M

j

= M

j

M

i

, i.e., the

M

i

's ommute.

Let

W

be the ve tor spa e spanned by the set

{M

1 , M

2 , . . . , M

q

}

, so the elements of

W

are

(n − 1) × (n − 1)

matri es. Suppose

k =

dim

W

. Then it follows from the indu tion hypothesis that

k ≤ ⌊(n − 1)

2 _{/4⌋ + 1}

. Assume, without loss of generality, that

W

is spanned by the linearly independent set of matri es

{M

1 , M

2 , . . . , M

k

}.

Then for

i > k

ea h

M

i

is expressible asa linear ombination

M

i

=

P

k

j=1

c

i,j

M

j

with s alars

c

i,1

, . . . , c

i,k

in

F.

Now, for

i > k

, let

B

i

= A

i

−

P

k

j=1

c

i,j

A

j

.

Thenfor

i = k + 1, . . . , ⌊n

2 _{/4⌋ + 2}

,

B

i

= A

i

−

k

X

j=1

c

i,j

A

j

=







H

i

0 M

i







−







P

k

j=1

c

i,j

H

j

0 P

k

j=1

c

i,j

M

j







=







t

i

0

0 





=

t

i

0

where

t

i

= H

i

−

P

k

j=1

c

i,j

H

j

is a

1 × n

matrix. Now onsider

q

X

i=k+1

d

i

B

i

=

q

X

i=k+1

d

i

A

i

−

k

X

j=1

c

i,j

A

j

= 0.

Sin e the

A

i

's are linearly independent it follows that the s alars

d

i

= d

i

c

i,j

= 0

and so the

B

i

'sare linearlyindependent. Furthermore,if

q

X

i=k+1

d

i

t

i

= 0

then

q

X

i=k+1

d

i

t

₀

i

=

q

X

i=k+1

d

i

B

i

= 0

and thus, the s alars

d

i

= 0

, hen e the set

{t

k+1

, . . . , t

q

}

is linearly independent over

F

.

Now onsider an alternative partitioning of ea h of

A

1 , A

2 , . . . , A

q

,

q = ⌊n

2 _{/4⌋ + 2}

into submatri es

M

′

i

,

H

′

i

and a zero matrix of order

(n − 1) × (n − 1)

,

n × 1

and

1 × (n − 1)

respe tively, as follows:

A

i

=







M

′

i

0 0 · · · 0

H

_i

′







(16)

forall

i

. Again,sin e

A

i

A

j

= A

j

A

i

itfollowsasbeforethat

M

′

i

M

j

′

= M

j

′

M

i

′

,i.e., the

M

′

i

's ommute. Let

W

′

be the ve tor spa e spanned by the set

{M

′

1 , M

2 ′

, . . . , M

q

′

}

, so the ele-mentsof

W

′

are

(n − 1) × (n − 1)

matri es. Suppose

s =

dim

W

′

. Thenfromtheindu tion

hypothesisit follows that

s ≤ ⌊(n − 1)

2 _{/4⌋ + 1}

. Assume,again withoutlossof generality,

that

W

′

is spanned by the linearly independent set of matri es

{M

′

1 , M

2 ′

, . . . , M

s

′

}.

Then for

i > s

ea h

M

′

i

is expressible as a linear ombination

M

′

i

=

P

s

j=1

c

′

i,j

M

j

′

with s alars

c

′

i,1

, . . . , c

′

i,s

in

F.

Now, for

i > s

, let

B

′

i

= A

i

−

P

s

j=1

c

i,j

′

A

j

.

A similar argument asbefore shows that

B

_i

′

=

0 t

′

i

where

t

′

i

= H

i

′

−

P

s

j=1

c

′

i,j

H

j

′

is an

n × 1

matrix,

i = s + 1, . . . , ⌊n

2 _{/4⌋ + 2.}

The

B

i

's are linear ombinations of the

A

i

's, so the

B

i

's are in

V

. Similarly the

B

′

j

's are in

V

and hen e

B

i

B

′

j

= B

j

′

B

i

. From

B

i

B

j

′

=

t

i

0 0 t

′

j

=

0 t

i

t

′

j

0

and

B

_j

′

B

i

=

0 t

′

j

t

i

0 = 0

itfollowsthat

t

i

t

′

j

= 0

forevery

i = k + 1, . . . , q

and

j = s + 1, . . . , q

,where

q = ⌊n

2 _{/4⌋+ 2}

. Let

A =







t

k+1

t

k+2

. . .

t

q







=







a

k+1,1

a

k+1,2

· · · a

k+1,n−1

a

k+1,n

a

k+2,1

a

k+2,2

· · · a

k+2,n−1

a

k+2,n

. . . . . .

· · ·

. . . . . .

a

q,1

a

q,2

· · ·

a

q,n−1

a

q,n







where

q = ⌊n

2 _{/4⌋ + 2}

and

t

i

= (a

i,1

, a

i,2

, . . . , a

i,n

)

for

i = k + 1, . . . , q.

Sin e the

t

i

's are linearly independent it follows that

rank

A ≥ ⌊n

2 _{/4⌋ + 2 − (k + 1) + 1 = ⌊n}

2 _{/4⌋ − k + 2.}

And

t

i

t

′

j

= 0

implies

At

′

j

= 0

whi h implies that

t

′

j

is in the null spa e of

A

for all

j = s + 1, . . . , ⌊n

2 _{/4⌋ + 2}

. Furthermore, the

t

′

i

's are linearly independent,thus,

nullity

A ≥ ⌊n

2 _{/4⌋ + 2 − (s + 1) + 1 = ⌊n}

2 _{/4⌋ − s + 2.}

Now

n =

rank

A +

nullity

A

≥

n

2

4 − k + 2

+

n

2

4 − s + 2

= 2

n

2

4 + 4 − (k + s).

(17)

CHAPTER 2. Mirzakhani'sSimple Proof of S hur's Theorem 10 But

k + s ≤ (⌊

(n−1)

2

4 ⌋ + 1 + ⌊

(n−1)

2

4 ⌋ + 1)

,so

n ≥ 2

n

2

4 + 4 −

(n − 1)

2

4 + 1 +

(n − 1)

2

4 + 1

= 2

n

2

4 −

(n − 1)

2

4 + 1

.

If

n

iseven, say

n = 2m, m ∈ Z

+

_,

then

n ≥ 2

(2m)

2

4 −

(2m − 1)

2

4 + 1

= 2

m

2 −

(m

2 − m) +

1

4 + 1

= 2(m

2 − m

2 + m + 1)

= 2m + 2

= n + 2.

If

n

isodd, say

n = 2m + 1, m ∈ Z

+

_,

then

n ≥ 2

(2m + 1)

2

4 −

(2m)

2

4 + 1

= 2

(m

2 _{+ m) +}

1

4 −

m

2 + 1

= 2(m

2 + m − m

2 + 1)

= 2m + 2

= n + 1.

Wehavethusarrived ata ontradi tion inboth ases. Thus, the assumptionthat

F

n

has more than

⌊n

2 _{/4⌋ + 1}

linearly independent matri es leads to a ontradi tion. Therefore

F

n

has atmost

⌊n

2 _{/4⌋ + 1}

linearly independent matri es.

Thefollowing ommutativesubalgebraof

M

n

(F )

givenin[9℄isanexampleofasubalgebra ontaining

⌊n

2 _{/4⌋ + 1}

linearly independent matri es. The example istrue for allpositive

integer

n

. This shows that the upper bound

⌊n

2 _{/4⌋ + 1}

annot be lowered. We look at

the ase when

n = 5

for illustrativepurposes. Example 15. The subalgebra

F

n

= {aI + a

i,j

E

i,j

| 1 ≤ i ≤ ⌊n/2⌋, ⌊n/2⌋ + 1 ≤ j ≤ n, a, a

i,j

∈ F }

of

M

n

(F )

,

F

aeld,isa ommutativesubalgebrawhi hhas

⌊n

2 _/4⌋+1

linearlyindependent

matri es.

We show that

F

n

is indeed ommutative. Sin e, by denition of

F

n

,

i

is never equal to

(18)

bI + b

i,j

E

i,j

in

F

n

we have

(aI + a

i,j

E

i,j

)(bI + b

i,j

E

i,j

) = abI + ab

i,j

E

i,j

+ a

i,j

bE

i,j

+ a

i,j

b

i,j

E

i,j

E

i,j

= abI + ab

i,j

E

i,j

+ a

i,j

bE

i,j

and

(bI + b

i,j

E

i,j

)(aI + a

i,j

E

i,j

) = baI + ba

i,j

E

i,j

+ b

i,j

aE

i,j

+ b

i,j

a

i,j

E

i,j

E

i,j

= baI + ba

i,j

E

i,j

+ b

i,j

aE

i,j

.

Sin e

F

isa eld, itthen follows that elements of

F

n

ommute.

Wenoti ethat every element

aI + a

i,j

E

i,j

of

F

n

an be expressed asa linear ombination

aI + a

i,j

E

i,j

= aI + a

1,⌊n/2⌋+1

E

1,⌊n/2⌋+1

+ . . . + a

1,n

E

1,n

+ a

2,⌊n/2⌋+1

E

2,⌊n/2⌋+1

+ . . . + a

2,n

E

2,n

. . .

+ a

⌊n/2⌋,⌊n/2⌋+1

E

⌊n/2⌋,⌊n/2⌋+1

+ . . . + a

⌊n/2⌋,n

E

⌊n/2⌋,n

and alsothe set

W = {I, E

i,j

| 1 ≤ i ≤ ⌊n/2⌋, ⌊n/2⌋ + 1 ≤ j ≤ n}

is linearly independent. Thus,

W

is abasis for

F

n

. The numberof elementsin

W

is

n

2 n −

n

2 + 1.

Now, if

n

is even, say

n = 2m

,

m ∈ Z

+

_,

then

n

2 n −

n

2 + 1 =

2m

2 2m −

2m

2 + 1

= ⌊m⌋(2m − ⌊m⌋) + 1

= m(2m − m) + 1

= m

2 + 1

=

(2m)

2

4 + 1

=

n

2

4 + 1.

(19)

CHAPTER 2. Mirzakhani'sSimple Proof of S hur's Theorem 12 If

n

isodd, say

n = 2m + 1, m ∈ Z

+

,then

n

2 n −

n

2 + 1 =

2m + 1

2 2m + 1 −

2m + 1

2 + 1

=

m +

1

2 2m + 1 −

m +

1

2 + 1

= m(2m + 1 − m) + 1

= m

2 + m + 1

=

m

2 + m +

1

4 + 1

=

(2m + 1)

2

4 + 1

=

n

2

4 + 1.

Example 16. In parti ular, if

n = 5

and

F

is aeld, then

F

5 = {aI + a

i,j

E

i,j

| 1 ≤ i ≤ ⌊5/2⌋, ⌊5/2⌋ + 1 ≤ j ≤ 5, a, a

i,j

∈ F }

= {aI + a

i,j

E

i,j

| 1 ≤ i ≤ 2, 3 ≤ j ≤ 5, a, a

i,j

∈ F }.

Let

A, B ∈ F

5

su h that

A =







a 0 a

1,3

a

1,4

a

1,5

0 a a

2,3

a

2,4

a

2,5

0 0

a

0

0 0 0

0 a

0 0 0

0

0 a







= aI + A

1,2

and

B =







b 0 b

1,3

b

1,4

b

1,5

0 b b

2,3

b

2,4

b

2,5

0 0

b

0

0 0 0

0 b

0 0 0

0

0 b







= bI + B

1,2

where

A

1,2

=







0 0 a

1,3

a

1,4

a

1,5

0 0 a

2,3

a

2,4

a

2,5

0 0

0

0 0 0

0

0 0 0

0

0 





and

B

1,2

=







0 0 b

1,3

b

1,4

b

1,5

0 0 b

2,3

b

2,4

b

2,5

0 0

0

0 0 0

0

0 0 0

0

0 





.

Wesee that

A

1,2

B

1,2

= B

1,2

A

1,2

= 0

and thus,

AB = (aI + A

1,2

)(bI + B

1,2

)

(20)

and

BA = (bI + B

1,2

)(aI + A

1,2

)

= baI + bA

1,2

+ aB

1,2

.

Therefore

AB = BA

indi atingthat the elements of

F

5

are indeed mutually ommuting. It is lear that

{I

5 , E

1,3

, E

1,4

, E

1,5

, E

2,3

, E

2,4

, E

2,5

}

isalinearlyindependentset ofmatri esin

F

5 .

Moreover, notethatthis sethas

7

elements and

⌊5

2 _{/4⌋ + 1 = 7.}

Weobservethat inthe subalgebra

F

n

inExample15 the number

⌊n

2 _{/4⌋ + 1}

isalsoequal

(21)

Chapter 3

Cayley-Hamilton Theorem

Inthis hapter wegiveadetailedproof ofthe Cayley-HamiltonTheoremfor anarbitrary

matrixin

M

n

(R),

where

R

denotestheeldofrealnumbers. Wefurtherdis ussinbriefthe tra eidentity fora

2 × 2

matrixover

R

whi h arisesfromthe Cayley-HamiltonTheorem.

Consider an

n × n

matrix

A ∈ M

n

(R)

given by

A = (a

i,j

) =







a

1,1

a

1,2

· · · a

1,n

a

2,1

a

2,2

· · · a

2,n

·

· · · ·

·

· · · ·

·

· · · ·

· a

n,1

a

n,2

· · · a

n,n







.

(3.1)

The followingdenitions are given in relationtothe general

n × n

matrix

A ∈ M

n

(R)

in Equation 3.1.

Denition 17. (Nering E.D. [10℄)

The determinant of the matrix

A = (a

i,j

)

is dened to be the s alar

det(A) = |a

i,j

|

omputed a ording tothe rule

det(A) = |a

i,j

| =

X

π

(

sgn

π)a

1,π(1)

a

2,π(2)

· · · a

n,π(n)

,

where the sum is taken overall permutationsof the elements of the set

S = {1, . . . , n}.

Wenote the following with regardto the above denition:

(i)"sgn

π

" means the "the sign of

π

".

(ii)A permutation

π

of a set

S

is dened to bea one to one mappingof

S

ontoitself. (iii)The element whi hthe permutation

π

asso iates with

i

is denoted by

π(i).

(iv) sgn

π = +1

if

π

is aneven permutation. (v) sgn

π = −1

if

π

isan odd permutation.

(22)

Example 18. Consider a

3 × 3

matrix

B ∈ M

3 (R)

given by

B =





b

1,1

b

1,2

b

1,3

b

2,1

b

2,2

b

2,3

b

3,1

b

3,2

b

3,3





.

(3.2) The determinant of

B

is

det(B) =

X

π

(

sgn

π)b

1,π(1)

b

2,π(2)

b

3,π(3)

= b

1,1

b

2,2

b

3,3

+ b

1,2

b

2,3

b

3,1

+ b

1,3

b

2,1

b

3,2

− b

1,2

b

2,1

b

3,3

− b

1,3

b

2,2

b

3,1

− b

1,1

b

2,3

b

3,2

= b

1,1

(b

2,2

b

3,3

− b

2,3

b

3,2

) + b

1,2

(b

2,3

b

3,1

− b

2,1

b

3,3

) + b

1,3

(b

2,1

b

3,2

− b

2,2

b

3,1

).

Thus,

det(B) = b

1,1

B

1,1

+ b

1,2

B

1,2

+ b

1,3

B

1,3

(3.3) whereea h

B

i,j

isadeterminantofamatrix obtainedfrom

B

bydeletingthe

ith

rowand

jth

olumnof

B.

That is,

B

1,1

=

b

2,2

b

2,3

b

3,2

b

3,3

, B

1,2

= −

b

2,1

b

2,3

b

3,1

b

3,3

, B

1,3

=

b

2,1

b

2,2

b

3,1

b

3,2

.

Remark 19. The determinantof the

n × n

matrix

A

in(3.1) an be writtenin the form

det(A) = a

i,j

A

i,j

+ (

terms whi hdo not ontain

a

i,j

as a fa tor

).

Example18andRemark19leadustothefollowingdenition,(seeJainandGunawardena

[5℄).

Denition 20. The ofa torof any entry

a

p,q

of an

n × n

matrix

A = (a

i,j

)

isdened to be

A

p,q

= (−1)

p+q

·det(

the

(n−1)×(n−1)

matrix obtained by deletingthe

p th

row and

q th

olumn

).

InRemark19thes alar

A

i,j

isthe ofa toroftheentry

a

i,j

andinExample18,

B

1,1

, B

1,2

, B

1,3

are ofa tors of

b

1,1

, b

1,2

, b

1,3

, respe tively.

Example 21. Usingthe matrix

B ∈ M

3 (R)

in Example 18, we have

b

2,1

B

1,1

+ b

2,2

B

1,2

+ b

2,3

B

1,3

(3.4)

= b

2,1

(b

2,2

b

3,3

− b

2,3

b

3,2

) + b

2,2

(b

2,3

b

3,1

− b

2,1

b

3,3

) + b

2,3

(b

2,1

b

3,2

− b

2,2

b

3,1

).

(3.5) Multiplying and grouping onthe right hand side of Equation 3.5gives

b

2,1

B

1,1

+ b

2,2

B

1,2

+ b

2,3

B

1,3

= b

2,1

b

2,2

b

3,3

− b

2,2

b

2,1

b

3,3

+ b

2,3

b

2,1

b

3,2

− b

2,1

b

2,3

b

3,2

+ b

2,2

b

2,3

b

3,1

− b

2,3

b

2,2

b

3,1

= 0.

(23)

CHAPTER 3. Cayley-Hamilton Theorem 16

Remark 22. In general

n

X

j=1

a

i,j

A

k,j

= a

i,1

A

k,1

+ a

i,2

A

k,2

+ . . . + a

i,n

A

k,n

= 0

whenever

i 6= k.

Denition 23. The transpose of the matrix

A = (a

i,j

)

is the matrix

A

T

whose element

a

i,j

appearing in row

i

and olumn

j

is the element

a

j,i

appearingin row

j

and olumn

i

of

A.

Againin relationto (3.1)and Denition20,the matrix

(A

i,j

) =







A

1,1

A

1,2

· · · A

1,n

A

2,1

A

2,2

· · · A

2,n

·

· · · ·

·

· · · ·

·

· · · ·

· A

n,1

A

n,2

· · · A

n,n







whoseentries are ofa torsof allentries of an

n × n

matrix

A = (a

i,j

)

is alled a ofa tor matrix.

Denition 24. The transpose

(A

j,i

) =







A

1,1

A

2,1

· · · A

n,1

A

1,2

A

2,2

· · · A

n,2

·

· · · ·

·

· · · ·

·

· · · ·

· A

1,n

A

2,n

· · · A

n,n







of the ofa tor matrix

(A

i,j

)

is alled the adjointof matrix

A

and itis denoted by

adj A.

Lemma 25. The produ t of any matrix

A ∈ M

n

(R)

and its adjoint

adj A

is equal to

(det(A))I,

i.e.,

A(adj A) = (det(A))I

where

I

is the identity matrix in

M

n

(R)

. Lemma26. Let

A

0 +A

1 λ+. . .+A

m

λ

m

_{= 0}

forall

|λ|

su iently large,where

A

i

∈ M

n

(R)

for all

i

and where

R

denotes the eld of real numbers. Then ea h

A

i

= 0.

Proof. Multiplying

A

0 + A

1 λ + . . . + A

m

λ

m

= 0

(3.6) by

1 λ

m

yields

A

0

1 λ

m

+ A

1

1 λ

m−1

+ . . . + A

m−1

1 λ

+ A

m

= 0.

(3.7)

(24)

Letting

|λ| → ∞

in(3.7) yields

A

m

= 0

and thus(3.7) be omes

A

0

1 λ

m

+ A

1

1 λ

m−1

+ . . . + A

m−2

1 λ

2 + A

m−1

1 λ

1 = 0.

(3.8) Multiplying (3.8) by

λ

yields

A

0

1 λ

m−1

+ A

1

1 λ

m−2

+ . . . + A

m−2

1 λ

+ A

m−1

= 0.

(3.9)

Letting

|λ| → ∞

in(3.9) yields

A

m−1

= 0

and thus (3.9) be omes

A

0

1 λ

m−1

+ A

1

1 λ

m−2

+ . . . + A

m−2

1 λ

1 = 0.

(3.10)

Continuing tomultiplyby

λ

and letting

λ → ∞

resultseventuallyin

A

0 λ

+ A

1 = 0.

(3.11)

Finally,againletting

λ → ∞

gives

A

1 = 0

and so(3.6)impliesthat

A

0 = 0.

Hen eallthe

A

i

's are zero.

Corollary 27. Let

A

i

and

B

i

be

n × n

matri es over

R

and suppose that

A

0 + A

1 λ + . . . + A

m

λ

m

= B

0 + B

1 λ + . . . + B

m

λ

m

for all

|λ|

large enough. Then

A

i

= B

i

for all

i

.

Proof. Sin e

A

0 − B

0 + (A

1 − B

1 )λ + . . . + (A

m

− B

m

)λ

m

= 0,

Lemma 26implies that

A

i

− B

i

= 0

for all

i

. Hen e

A

i

= B

i

for all

i

.

ThefollowingtheoremisthewellknownCayley-HamiltonTheoremanditstatesthatany

n × n

matrix

A

isa solution of its hara teristi polynomial

p(λ).

Theorem 28. Let

A

be an

n × n

matrix over

R

and let

p(λ) = det(A − λI)

be the hara teristi polynomial of

A

. Then

p(A) = 0.

Proof. Consider a matrix

A ∈ M

n

(R)

su hthat

A = (a

i,j

).

Then

A − λI =







a

1,1

− λ

a

1,2

· · ·

a

1,n

a

2,1

a

2,2

− λ · · ·

a

2,n

· · ·

a

n,1

a

n,2

· · · a

n,n

− λ







and the determinantof

A − λI

isa polynomial in

λ

of degree

n

, say,

(25)

for some

k

0 , k

1 , . . . , k

n

∈ R.

Now the ofa tor

A

1,1

=

a

2,2

− λ · · ·

a

2,n

· · ·

a

n,2

· · · a

n,n

− λ

is apolynomialin

λ

of degree

n − 1

. In fa tea h ofa tor

A

i,j

is apolynomialin

λ

of at most

(n−1)

thdegree. Hen etheentriesoftheadjointmatrix

adj(A−λI)

arepolynomials in

λ

of atmost

(n − 1)

thdegree. Thus,

adj(A − λI) = B

n−1

λ

n−1

+ B

n−2

λ

n−2

+ . . . + B

1 λ + B

0

where ea h

B

i

is an

n × n

matrix with entries from

R

. Now

(A − λI) adj(A − λI)

= (A − λI)(B

n−1

λ

n−1

+ B

n−2

λ

n−2

+ . . . + B

1 λ + B

0 )

= AB

n−1

λ

n−1

− IB

n−1

λ

n

+ AB

n−2

λ

n−2

− IB

n−2

λ

n−1

+ . . . + AB

1 λ − IB

1 λ

2 + AB

0 − IB

0 λ,

i.e.,

(A − λI) adj(A − λI) = C

n

λ

n

+ C

n−1

λ

n−1

+ C

n−2

λ

n−2

+ . . . + C

1 λ + C

0

(3.13) where

C

n

= −IB

n−1

C

n−1

= AB

n−1

− IB

n−2

C

n−2

= AB

n−2

− IB

n−3

. . .

C

2 = AB

2 − IB

1 C

1 = AB

1 − IB

0 C

0 = AB

0 .

Multiplying the above equationsfrom the left by

A

n

, A

n−1

, A

n−2

, . . . , A

2 , A, I

respe tively, yields

A

n

C

n

= A

n

(−IB

n−1

)

A

n−1

C

n−1

= A

n−1

(AB

n−1

− IB

n−2

)

A

n−2

C

n−2

= A

n−2

(AB

n−2

− IB

n−3

)

. . .

A

2 C

2 = A

2 (AB

2 − IB

1 )

AC

1 = A(AB

1 − IB

0 )

IC

0 = IAB

0 .

(26)

Addingthe aboveequations yields

A

n

C

n

+ A

n−1

C

n−1

+ A

n−2

C

n−2

+ . . . + A

2 C

2 + AC

1 + C

0 = −A

n

_B

n−1

+ A

n

B

n−1

− A

n−1

B

n−2

+ A

n−1

B

n−2

− A

n−2

B

n−3

+ . . . + A

3 B

2 −A

2 B

1 + A

2 B

1 − AB

0 + AB

0 = 0.

That is,

A

n

C

n

+ A

n−1

C

n−1

+ A

n−2

C

n−2

+ . . . + A

2 C

2 + AC

1 + C

0 = 0.

(3.14) Now, itfollows fromLemma 25 and (3.12)that

(A − λI)

adj

(A − λI) = I(

det

(A − λI)) = I(k

n

λ

n

_{+ k}

n−1

λ

n−1

+ . . . + k

1 λ + k

0 ).

(3.15)

Substituting (3.13)into (3.15) gives

C

n

λ

n

+ C

n−1

λ

n−1

+ C

n−2

λ

n−2

+ . . . + C

1 λ + C

0 = I(k

n

λ

n

+ k

n−1

λ

n−1

+ . . . + k

1 λ + k

0 ).

By Corollary27 wehave

C

n

= Ik

n

, C

n−1

= Ik

n−1

, C

n−2

= Ik

n−2

, · · · , C

2 = Ik

2 , C

1 = Ik

1 , C

0 = Ik

0 .

Substituting the

C

i

by

Ik

i

in(3.14) we get

k

n

A

n

+ k

n−1

A

n−1

+ k

n−2

A

n−2

+ . . . + k

2 A

2 + k

1 A + k

0 I = 0.

Consider a

2 × 2

matrix

A =

a b

c d

over

R

with hara teristi polynomial

p(λ) = λ

2 − (a + d)λ + ad − bc

= λ

2 − tr(A)λ + det(A).

The Cayley-Hamilton Theorem asserts that

p(A) = A

2 _{− tr(A)A + I det(A) = 0,}

(3.16)

where

I

is the identity matrix in

M

2 (R).

Sin e

tr(I) = 2

, taking the tra e of (3.16), it follows that

tr(A

2 ) − tr

2 (A) + 2 det(A) = 0.

(3.17)

Thus,

det(A) =

1

2 tr

2 _{(A) − tr(A}

2 _).

(3.18)

Substituting (3.18)into (3.16) gives

A

2 − tr(A)A +

1

2 tr

2 _{(A) − tr(A}

2 _{)I = 0.}

(27)

The identity in(3.19) isknown as the Cayley-Hamiltontra e identity for

2 × 2

matri es. Weshall see in Chapter

5

that (3.19) is not ne essarilyzero when we deal with matri es over anon ommutativering.

We note that, although we have dealt with the Cayley-Hamilton Theorem for square

matri esover the eld of real numbers, there exist several proofs of this theorem for

ma-tri esoverthe eld

C

of omplexnumbers andarbitrary ommutativerings. Forinstan e, Zhang[14℄provided a prooffor square matri esoverthe eld

C

of omplexnumbers and Straubing [12℄ provided a proof for a square matrix over anarbitrary ommutativering.

Furthermore,toeverylineartransformationofave torspa eofdimension

n

overaeld

F

there orresponds an

n × n

matrix over

F

and onversely,toeverysu h matrixthere or-responds a linear transformation of the ve tor spa e. A ordingly, the Cayley-Hamilton

Theorem hold for linear transformations and Hungerford [3℄ provides a proof for linear

(28)

Chapter 4

Commutativity and Lie nilpoten y in

the Matrix Algebra

U

∗

n

(R)

Inthis hapterwedisse tthepaper[8℄byMeyer, etal. Weexplorethering

U

∗

n

(R)

,in par-ti ularthe rings

U

∗

3 (R)

and

U

∗

4 (R)

,inrelationtothe polynomial identities

[[x, y], z] = 0,

[x, y][u, v] = 0

and

[[x, y], [u, v]] = 0.

Wedis ussboth ases whenaring

R

is ommutative andwhen

R

isnot ommutative. Itisshown inTheorem44that

U

∗

n

(R)

isLienilpotentof index

n − 1

. A ring satisfying anidentity

[[x, y], [u, v]] = 0

is alled Lie solvable of index

2

. In [8℄, Meyer, etal,giveanexample of a Liesolvable ring of index

2

whi h wedis uss in Corollary37.

4.1 The ring

U

∗

3 (R)

Denition 29. Let

R

be an arbitrary ring (not ne essarily ommutative) with unity

1

, and let

U

_n

∗

(R) =

















a a

1,2

· · ·

a

1,n

0 a

. . . . . . . . . . . . . . .

a

n−1,n

0 · · ·

0 a







: a, a

i,j

∈ R











be the subring of

M

n

(R)

of uppertriangularmatri es with equaldiagonalentries. Example 30. Consider the elements