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1 Linear Transformations

We will study mainly nite-dimensional vector spaces over an arbitrary eld F|i.e. vector spaces with a basis. (Recall that the dimension of a vector spaceV(dimV) is the number of elements in a basis ofV.)

DEFINITION 1.1

(Linear transformation) Given vector spacesUandV,T:U7!Vis a linear transformation (LT) if

T(u+v) =T(u) +T(v)

for all;2F, andu;v2U. ThenT(u+v) =T(u)+T(v); T(u) =T(u) and T nX k=1 kuk! =nX k=1 kT(uk):

EXAMPLES 1.1

Consider the linear transformation

T=TA:Vn(F)7!Vm(F)

whereA= [aij] ismn, dened byTA(X) =AX. Note thatVn(F) = the set of alln-dimensional column vectors2 6 4x 1... x n3 7 5of

F|sometimes writtenFn.

Note that ifT:Vn(F)7!Vm(F) is a linear transformation, thenT=TA, whereA= [T(E1)jjT(En)] and E 1=2 6 6641
0 03 7

775;:::;En=2

6 6640
0 13 7 775
Note: v2Vn(F); v=2 6 4x 1... x n3 7

5=x1E1++xnEn

1 IfVis a vector space of all innitely dierentiable functions onR, then

T(f) =a0Dnf+a1Dn1f++an1Df+anf

denes a linear transformationT:V7!V. The set offsuch thatT(f) = 0 (i.e. the kernel ofT) is important. LetT:U7!Vbe a linear transformation. Then we have the following denition:

DEFINITIONS 1.1

(Kernel of a linear transformation)

KerT=fu2UjT(u) = 0g

(Image ofT)

ImT=fv2Vj 9u2Usuch thatT(u) =vg

Note: KerTis a subspace ofU. Recall thatWis a subspace ofUif

1. 02W,

2.Wis closed under addition, and

3.Wis closed under scalar multiplication.

PROOF. that KerTis a subspace ofU:

1.T(0) + 0 =T(0) =T(0 + 0) =T(0) +T(0). ThusT(0) = 0, so

02KerT.

2. Letu;v2KerT; thenT(u) = 0 andT(v) = 0. SoT(u+v) =

T(u) +T(v) = 0 + 0 = 0 andu+v2KerT.

3. Letu2KerTand2F. ThenT(u) =T(u) =0 = 0. So

u2KerT.

EXAMPLE 1.1

KerTA=N(A);the null space of A

=fX2Vn(F)jAX= 0g andImTA=C(A);the column space of A =hA1;:::;Ani 2 Generally, ifU=hu1;:::;uni, then ImT=hT(u1);:::;T(un)i. Note:Even ifu1;:::;unform a basis forU,T(u1);:::;T(un) may not form a basis for ImT. I.e. it may happen thatT(u1);:::;T(un) are linearly dependent.

1.1 Rank + Nullity Theorems (for Linear Maps)

THEOREM 1.1 (General rank + nullity theorem)

IfT:U7!Vis a linear transformation then

rankT+ nullityT= dimU:

PROOF.

1. KerT=f0g.

Then nullityT= 0.

We rst show that the vectorsT(u1);:::;T(un), whereu1;:::;unare a basis forU, are LI (linearly independent):

Supposex1T(u1) ++xnT(un) = 0 wherex1;:::;xn2F.

Then

T(x1u1++xnun) = 0 (by linearity)

x

1u1++xnun= 0 (since KerT=f0g)

x

1= 0;:::;xn= 0 (sinceuiare LI)

Hence ImT=hT(u1);:::;T(un)iso

rankT+ nullityT= dim ImT+ 0 =n= dimV:

2. KerT=U.

So nullityT= dimU.

Hence ImT=f0g )rankT= 0

)rankT+ nullityT= 0 + dimU = dimU:

3. 0 Letu1;:::;urbe a basis for KerTandn= dimU, sor= nullityT andr < n. Extend the basisu1;:::;urto form a basisu1;:::;ur;ur+1;:::;unof 3 U(refer to last year's notes to show that this can be done).

ThenT(ur+1);:::;T(un) span ImT. For

ImT=hT(u1);:::;T(ur);T(ur+1);:::;T(un)i

=h0;:::;0;T(ur+1);:::;T(un)i =hT(ur+1);:::;T(un)i

So assume

x

1T(ur+1) ++xnrT(un) = 0

)T(x1ur+1++xnrun) = 0 )x1ur+1++xnrun2KerT )x1ur+1++xnrun=y1u1++yrur for somey1;:::;yr )(y1)u1++ (yr)ur+x1ur+1++xnrun= 0 and sinceu1;:::;unis a basis forU, all coecients vanish. Thus rankT+ nullityT= (nr) +r =n = dimU: We now apply this theorem to prove the following result:

THEOREM 1.2 (Dimension theorem for subspaces)

dim(U\V) + dim(U+V) = dimU+ dimV whereUandVare subspaces of a vector spaceW. (Recall thatU+V=fu+vju2U;v2Vg.)

For the proof we need the following denition:

DEFINITION 1.2

IfUandVare any two vector spaces, then the direct sum is

UV=f(u;v)ju2U;v2Vg

(i.e. the cartesian product ofUandV) made into a vector space by the component-wise denitions: 4

1.(u1;v1) + (u2;v2) = (u1+u2;v1+v2),

2.(u;v) = (u;v), and

3.(0;0)is an identity forUVand(u;v)is an additive inverse for

(u;v).

We need the following result:

THEOREM 1.3

dim(UV) = dimU+ dimV

PROOF.

Case 1:U=f0g

Case 2:V=f0g

Proof of cases 1 and 2 are left as an exercise.

Case 3:U6=f0gandV6=f0g

Letu1;:::;umbe a basis forU, and

v

1;:::;vnbe a basis forV.

We assert that (u1;0);:::;(um;0);(0;v1);:::;(0;vn) form a basis forUV.

Firstly, spanning:

Let (u;v)2UV, sayu=x1u1++xmumandv=y1v1++ynvn.

Then (u;v) = (u;0) + (0;v) = (x1u1++xmum;0) + (0;y1v1++ynvn) =x1(u1;0) ++xm(um;0) +y1(0;v1) ++yn(0;vn) Secondly, independence: assumex1(u1;0)++xm(um;0)+y1(0;v1)+ +yn(0;vn) = (0;0). Then (x1u1++xmum;y1v1++ynvn) = 0 )x1u1++xmum= 0 andy1v1++ynvn= 0 )xi= 0;8i andyi= 0;8i 5 Hence the assertion is true and the result follows.

PROOF.

LetT:UV7!U+VwhereUandVare subspaces of someW, such thatT(u;v) =u+v.

Thus ImT=U+V, and

KerT=f(u;v)ju2U;v2V;andu+v= 0g

=f(t;t)jt2U\Vg

Clearly then, dim KerT= dim(U\V)1and so

rankT+ nullityT= dim(UV) )dim(U+V) + dim(U\V) = dimU+ dimV:

1.2 Matrix of a Linear Transformation

DEFINITION 1.3

LetT:U7!Vbe a LT with bases:u1;:::;unand

:v1;:::;vmfor

UandVrespectively.

Then

T(uj) =a

1jv1 a 2jv2 a mjvmfor some a 1j... a mj2F:

Themnmatrix

A= [aij]

is called the matrix ofTrelative to the basesand and is also written

A= [T]

Note:Thej-th column ofAis the co-ordinate vector ofT(uj), where u jis thej-th vector of the basis.

Also ifu=x1u1++xnun, the co-ordinate vector2

6 4x 1... x n3 7

5is denoted by

[u].1True ifU\V=f0g; if not, letS= KerTandu1;:::;urbe a basis forU\V. Then (u1;u1);:::;(ur;ur) form a basis for S and hence dim KerT= dimS. 6

EXAMPLE 1.2

LetA=a b

c d

2M22(F)and letT:M22(F)7!M22(F)be

dened by

T(X) =AXXA:

ThenTis linear2, andKerTconsists of all22matricesAwhereAX= XA.

Taketo be the basisE11,E12,E21, andE22, dened by

E

11=1 0

0 0 ;E

12=0 1

0 0 ;E

21=0 0

1 0 ;E

22=0 0

0 1 (so we can dene a matrix for the transformation, consider these henceforth to be column vectors of four elements).

Calculate[T]

=B:

T(E11) =AE11E11A

=a b c d 1 0 0 0 1 0 0 0 a b c d 0b c0 = 0E11bE12+cE21+ 0E22 and similar calculations for the image of other basis vectors show that B=2 6

640c b0

b ad0b c0dac

0c b03

7 75
Exercise:Prove thatrankB= 2ifAis not a scalar matrix (i.e. if

A6=tIn).

Later, we will show thatrankB= rankT. Hence

nullityT= 42 = 22

T(X+Y) =A(X+Y)(X+Y)A

=(AXXA) +(AYY A) =T(X) +T(Y) 7 Note:I2;A2KerTwhich has dimension2. Hence ifAis not a scalar matrix, sinceI2andAare LI they form a basis forKerT. Hence

AX=XA)X=I2+A:

DEFINITIONS 1.2

LetT1andT2be LT's mapping U to V.

ThenT1+T2:U7!Vis dened by

(T1+T2)(x) =T1(x) +T2(x);8x2U

ForTa LT and2F, deneT:U7!Vby

(T)(x) =T(x)8x2U

Now ...

[T1+T2] = [T1] + [T2] [T] =[T]

DEFINITION 1.4

Hom(U;V) =fTjT:U7!Vis a LTg:

Hom(U;V)is sometimes writtenL(U;V).

The zero transformation 0 :U7!Vis such that 0(x) = 0,8x.

IfT2Hom(U;V), then (T)2Hom(U;V) is dened by

(T)(x) =(T(x))8x2U:

Clearly, Hom(U;V) is a vector space.

Also [0] = 0 and [T] =[T] The following result reduces the computation ofT(u) to matrix multi- plication:

THEOREM 1.4

[T(u)] = [T] [u] 8

PROOF.

LetA= [T]

, whereis the basisu1;:::;un, is the basisv1;:::;vm, and

T(uj) =mX

i=1a ijvi:

Also let [u]=2

6 4x 1... x n3 7 5.

Thenu=Pn

j=1xjuj, so

T(u) =nX

j=1x jT(uj) nX j=1x jm X i=1a ijvi mX i=10 nX j=1a ijxj1 A vi )[T(u)] =2 6 4a

11x1++a1nxn...

a m1x1++amnxn3 7 5 =A[u]

DEFINITION 1.5

(Compositionof LTs)

IfT1:U7!VandT2:V7!Ware LTs, thenT2T1:U7!Wdened by

(T2T1)(x) =T2(T1(x))8x2U is a LT.

THEOREM 1.5

If, andare bases forU,VandW, then [T2T1] = [T2] [T1] 9

PROOF. Letu2U. Then

[T2T1(u)]= [T2T1] [u] and = [T2(T1(u))] = [T2] [T1(u)] Hence [T2T1] [u]= [T2] [T1] [u](1) (note that we can't just \cancel o" the [u]to obtain the desired result!) Finally, ifisu1;:::;un, note that [uj]=Ej(sinceuj= 0u1++

0uj1+ 1uj+ 0uj+1++ 0un) then for an appropriately sized matrixB,

BE j=Bj;thejth column ofB.

Then (1) shows that the matrices

[T2T1] and [T2] [T1] have their rst, second, ...,nth columns respectively equal.

EXAMPLE 1.3

IfAismnandBisnp, then

T

ATB=TAB:

DEFINITION 1.6

(theidentity transformation) LetUbe a vector space. Then the identity transformationIU:U7!U dened by I

U(x) =x8x2U

is a linear transformation, and [IU] =Inifn= dimU.

Also note thatIVn(F)=TIn.

THEOREM 1.6

LetT:U7!Vbe a LT. Then

I

VT=TIU=T:

10 Then T

ImTA=TImA=TA=TATAIn=TAIn

and consequently we have the familiar result I mA=A=AIn:

DEFINITION 1.7

(Invertible LTs)

LetT:U7!Vbe a LT.

If9S:V7!Usuch thatSis linear and satises

ST=IUandTS=IV

then we say thatTisinvertibleand thatSis aninverseofT.

Such inverses are unique and we thus denoteSbyT1.

Explicitly,

S(T(x)) =x8x2UandT(S(y)) =y8y2V

There is a corresponding denition of aninvertible matrix:A2Mmn(F) is called invertible if9B2Mnm(F)such that

AB=ImandBA=In

Evidently

THEOREM 1.7

T Ais invertible iAis invertible (i.e. ifA1exists). Then, (TA)1=TA1

THEOREM 1.8

Ifu1;:::;unis a basis forUandv1;:::;vnare vectors inV, then there is one and only one linear transformationT:U!Vsatisfying

T(u1) =v1;:::;T(un) =vn;

namelyT(x1u1++xnun) =x1v1++xnvn. (In words, a linear transformation is determined by its action on a basis.) 11

1.3 Isomorphisms

DEFINITION 1.8

A linear mapT:U7!Vis called anisomorphismifTis 1-1 and onto, i.e.

1.T(x) =T(y))x=y8x;y2U, and

2.ImT=V, that is, ifv2V,9u2Usuch thatT(u) =v.

Lemma:A linear mapTis 1-1 i KerT=f0g.

Proof:

1. ()) SupposeTis 1-1 and letx2KerT.

We haveT(x) = 0 =T(0), and sox= 0.

2. (() Assume KerT=f0gandT(x) =T(y) for somex;y2U.

Then

T(xy) =T(x)T(y) = 0

)xy2KerT )xy= 0)x=y

THEOREM 1.9

LetA2Mmn(F). ThenTA:Vn(F)!Vm(F)is

(a) onto:,dimC(A) =m,the rows ofAare LI; (b) 1{1:,dimN(A) = 0,rankA=n,the columns ofAare LI.

EXAMPLE 1.4

LetTA:Vn(F)7!Vn(F)withAinvertible; soTA(X) =AX.

We will show this to be an isomorphism.

1. LetX2KerTA, i.e.AX= 0. Then

A

1(AX) =A10

)InX= 0 )X= 0 )KerT=f0g ,Tis 1-1. 12

2. LetY2Vn(F): then,

T(A1Y) =A(A1Y)

=InY=Y soImTA=Vn(F)

THEOREM 1.10

IfTis an isomorphism betweenUandV, then

dimU= dimV

PROOF.

Letu1;:::;unbe a basis forU. Then

T(u1);:::;T(un)

is a basis forV(i.e.huii=UandhT(ui)i=V, withui,viindependent families), so dimU=n= dimV

THEOREM 1.11

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