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Linear Transformation notes

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Sean Uppal

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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.)
(Linear transformation)
Given vector spaces Uand V,T:U7→ Vis alinear transformation (LT)
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)
T n
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
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
, . . . , En=
=x1E1+· · · +xnEn
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
(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.
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
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
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.
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.
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

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1 Linear Transformations We will study mainly nite-dimensional vector spaces over an arbitrary eld Fi.e. vector spaces with a basis. (Recall that the dimension of a vector space V (dimV ) is the number of elements in a basis of V .) DEFINITION 1.1 (Linear transformation) Given vector spaces U and V , T : U V is a linear transformation (LT) if T(u + v) = T(u) + T(v) for all , F, and u,v U. Then T(u+v) = T(u)+T(v), T(u) = T(u) and n T u = T(u ). k k k k k=1 k=1 EXAMPLES 1.1 Consider the linear transformation T = TA: Vn(F) m (F) where A = [ij] is m n, dened Ay T (X) = AX. x1 Note that V (F) = the set of all n-dimensional column ector of n . xn Fsometimes written F . Note that if Tn: V (F)m V (F) is a linear transformationA then T = T , where A = [T(E )T(E )] and 1 n 1 0 0 . E = ,...,E = . 1 . n 0 0 1 Note: x1 v V (F), v = . = x E + + x E n . 1 1 n n xn 1 www.notesolution.comIf V is a vector space of all innitely dierentiable functions on R, then T(f) = a D f + a D n1f + + a Df + a f 0 1 n1 n denes a linear transformation T : V V . The set of f such that T(f) = 0 (i.e. the kernel of T) is important. Let T : U V be a linear transformation. Then we have the following denition: DEFINITIONS 1.1 (Kernel of a linear transformation) KerT = {u U T(u) = 0} (Image of T) ImT = {v V u U such that T(u) = v} Note: KerT is a subspace of U. Recall that W is a subspace of U if 1. 0 W, 2. W is closed under addition, and 3. W is closed under scalar multiplication. PROOF. that KerT is a subspace of U: 1. T(0) + 0 = T(0) = T(0 + 0) = T(0) + T(0). Thus T(0) = 0, so 0 KerT. 2. Let u,v KerT; then T(u) = 0 and T(v) = 0. So T(u + v) = T(u) + T(v) = 0 + 0 = 0 and u + v KerT. 3. Let u KerT and F. Then T(u) = T(u) = 0 = 0. So u KerT. EXAMPLE 1.1 KerT A = N(A), the null space of A = {X V (n) AX = 0} and ImT = C(A), the column space of A A = A ,1.,A n 2
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