Linear Transformation notes

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21 Jan 2011
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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, µF,and u, vU.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:
vVn(F),v=
x1
.
.
.
xn
=x1E1+· · · +xnEn
1
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If Vis avector space of all infinitely differentiable functions on R,then
T(f)=a0Dnf+a1Dn1f+· · · +an1Df+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={uU|T(u)=0}
(Image of T)
Im T={vV|uUsuchthat T(u)=v}
Note:Ker Tis asubspace of U.Recall that Wis asubspace of Uif
1. 0W,
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
0Ker T.
2. Let u, vKer T;then T(u)=0and T(v)=0. So T(u+v)=
T(u)+T(v)=0+0=0and u+vKer T.
3. Let uKer Tand λF.Then T(λu)=λT (u)=λ0=0. So
λu Ker T.
EXAMPLE 1.1
Ker TA=N(A),the null space of A
={XVn(F)|AX =0}
and Im TA=C(A),the column space of A
=hA1, . . . , Ani
2
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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, . . . , xnF.
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
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