Department

Mathematics

Course Code

MAT223H1

Professor

Sean Uppal

1Linear Transformations

Wewill study mainly ﬁnite-dimensional vector spaces over an arbitrary ﬁeld

F—i.e. vector spaces with abasis. (Recall that the dimension of avector

space V(dim V)is the number of elements in abasis of V.)

DEFINITION 1.1

(Linear transformation)

Given vector spaces Uand V,T:U7→ Vis alinear transformation (LT)

if

T(λu +µv)=λT (u)+µT (v)

for all λ, µ∈F,and u, v∈U.Then T(u+v)=T(u)+T(v),T(λu)=λT (u)

and

T n

X

k=1

λkuk!=

n

X

k=1

λkT(uk).

EXAMPLES 1.1

Consider the linear transformation

T=TA:Vn(F)7→ Vm(F)

where A=[aij ]is m×n,deﬁned byTA(X)=AX.

Note that Vn(F)=the set of all n-dimensional column vectors

x1

.

.

.

xn

of

F—sometimes written Fn.

Note that if T:Vn(F)7→ Vm(F)is alinear transformation, then T=TA,

where A=[T(E1)| · · · |T(En)] and

E1=

1

0

.

.

.

0

, . . . , En=

0

.

.

.

0

1

Note:

v∈Vn(F),v=

x1

.

.

.

xn

=x1E1+· · · +xnEn

1

www.notesolution.com

If Vis avector space of all inﬁnitely diﬀerentiable functions on R,then

T(f)=a0Dnf+a1Dn−1f+· · · +an−1Df+anf

deﬁnes alinear transformation T:V7→ V.

The set of fsuchthat T(f)=0(i.e. the kernel of T)is important.

Let T:U7→ Vbealinear transformation. Then wehavethe following

deﬁnition:

DEFINITIONS 1.1

(Kernel of alinear transformation)

Ker T={u∈U|T(u)=0}

(Image of T)

Im T={v∈V|∃u∈Usuchthat T(u)=v}

Note:Ker Tis asubspace of U.Recall that Wis asubspace of Uif

1. 0∈W,

2. Wis closed under addition, and

3. Wis closed under scalar multiplication.

PROOF. that Ker Tis asubspace of U:

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

0∈Ker T.

2. Let u, v∈Ker T;then T(u)=0and T(v)=0. So T(u+v)=

T(u)+T(v)=0+0=0and u+v∈Ker T.

3. Let u∈Ker Tand λ∈F.Then T(λu)=λT (u)=λ0=0. So

λu ∈Ker T.

EXAMPLE 1.1

Ker TA=N(A),the null space of A

={X∈Vn(F)|AX =0}

and Im TA=C(A),the column space of A

=hA∗1, . . . , A∗ni

2

www.notesolution.com

Generally,if U=hu1, . . . , uni,then Im T=hT(u1), . . . , T(un)i.

Note: Even if u1, . . . , unform abasis for U,T(u1), . . . , T(un)maynot

form abasis for Im T.I.e. it mayhappen that T(u1), . . . , T(un)are linearly

dependent.

1.1 Rank +NullityTheorems (for Linear Maps)

THEOREM 1.1 (General rank +nullitytheorem)

If T:U7→ Vis alinear transformation then

rank T+nullityT=dim U.

PROOF.

1. Ker T={0}.

Then nullityT=0.

Weﬁrst showthat the vectors T(u1), . . . , T(un), where u1, . . . , unare

abasis for U,are LI (linearly independent):

Suppose x1T(u1)+· · · +xnT(un)=0where x1, . . . , xn∈F.

Then

T(x1u1+· · · +xnun)=0(bylinearity)

x1u1+· · · +xnun=0(since Ker T={0})

x1=0, . . . , xn=0(since uiare LI)

Hence Im T=hT(u1), . . . , T(un)iso

rank T+nullityT=dim Im T+0=n=dim V.

2. Ker T=U.

So nullityT=dim U.

Hence Im T={0}⇒rank T=0

⇒rank T+nullityT=0+dim U

=dim U.

3. 0<nullityT<dim U.

Let u1, . . . , urbeabasis for Ker Tand n=dim U,so r=nullityT

and r<n.

Extend the basis u1, . . . , urto form abasis u1, . . . , ur,ur+1, . . . , unof

3

www.notesolution.com

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