4 Linear Transformation

Linear Transformations

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Definition : Let U and V be real vector spaces. A map T: U → V from U to V is called a linear map, or Linear transformation, if T satisfies the following conditions

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

for all

(ii) T(αu) =αT(u) for all u ∈ U

α and all real numbers Rajiv Kumar Math II

u , v ∈U

Example (i) Reflection about x-axis (ii) Reflection about y-axis (iii)Projection on x-axis (iv)Projection on y-axis

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T(x,y)= (x,-y) T(x,y)= (-x,y) T(x,y)=(x,0) T(x,y)=(0,y)

Example: Let a map T: V2→V3 be defined by T(x1, x2) = (x1, x2 ,x1 x2). Check whether T is linear.

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T (α(x1,x2)) = T(αx1, αx2) = (αx1, αx2 , αx1 αx2 ) = (αx1, αx2 , α2x1x2) = α(x1, x2 αx1x2 ) ≠α(x1, x2 ,x1 x2). Hence T is not linear. Rajiv Kumar Math II

Example: Let C[-1, 1] be the vector space of real valued continuous functions defined on interval [-1, 1], T(f(x)) = x f ′(x) for all f(x) ∈C[-1, 1]. show that T is a linear map.

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Let f(x), g(x) ∈C[-1,1] T( f (x) + g(x)) = x (f(x)+g(x)' = x(f '(x)+g'(x) ) = x f'(x) +x g'(x) =T( f (x) )+ T(g(x)) T (αf(x)) = x [αf (x)]' =x(αf '(x))= α T (f(x))

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Theorem: Let T: U→V be a linear map. then • T(0U)=0V, • T(-u) = -T(u), and • T(α1u1+…+αnun)= α1T(u1)+…+ α nT(un). In other words, a linear map T transforms the 0U into the 0V and the negative of every u∈U into the negative of T(u) in V. Rajiv Kumar Math II

Proof of (i) As T is linear

T (αu ) = αT ( u ) If

α = 0, ⇒ T ( 0u ) = 0T ( u ) ⇒ T ( 0u ) = 0v

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induction on ‘n’, starting from the fact that T(αu) = αT(u) and using the property T(αu1+βu2)=T(αu1)+T(βu2)=αT(u1)+ βT(u2).

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In view of (iii), we get a technique of defining a linear transformation T on a finite-dimensional vector space U. Suppose B ={u1,u2,…..,un}is a basis for U. Then any vector u∈U can be expressed uniquely in the form u = α1u1+ α2u2+…+ αnun. Rajiv Kumar Math II

So, if T:U→V is a linear map, then

T(u) =T(α1u1+α2u2+…+ αnun) = α1T(u1)+ α2T(u2)+…+ αnT(un). Thus, T(u) is known as soon as T(u1), T(u2),…T(un) are known,

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Theorem: A linear transformation T is completely determined by its values on the elements of a basis. Precisely, if B={u1,u2,… un} is a basis for U and v1, v2, ….vn be n vectors (not necessarily distinct) in V Rajiv Kumar Math II

then

there

exists

a

unique

transformation T:U→V ,Such that T(ui)=vi

for 1 ≤i ≤n

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linear

Q.2 Determine whether there exists a linear map satisfying conditions. If it does exist give general formula. • T : V2→V2 T(1,1)=(3,0) ,T(-1,1)= (0,1) (ii) T : V2→V2 T(0,1)= (3,4) , T(3,1)=(2,2) & T(3,2)=(5,7) (iii) T : P3→P3 , such that T(1+x)=1+x T(2+x)= x+3x3 & T(x2)=0 (iv) T : V3→V3 T(0,1,2)= (3,1,2) , T(1,1,1)=(2,2,2) Rajiv Kumar Math II

What is general idea in doing such a problem? (ii) If set of elements on which transformation T is defined is independent then transformation T exist. (ii) If set of elements on which transformation T is defined is dependent then transformation T may or may not exist.

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T : V2→V2 , T(1,1)=(3,0) ,T(-1,1)= (0,1) {(1,1), (-1,1)} is linearly independent. , hence

such a linear transformation exist . {(1,1), (-1,1)} is basis of V2

To find linear transformation we express (x,y) as linear combination of {(1,1), (-1,1)}. Rajiv Kumar Math II

(x,y) = α(1,1)+β(-1,1)= (α-β,α+β) ⇒ α-β=x & α+β=y Solving we get α=(x+y)/2 & β = (y-x)/2 (x,y) = ((x+y)/2) (1,1)+ ((y-x)/2) (-1,1) T(x,y) = ((x+y)/2) T(1,1)+ ((y-x)/2) T(-1,1),

T(x,y) = ((x+y)/2) (3,0)+ ((y-x)/2) (0,1) = ((3x+3y)/2, (y-x)/2) Rajiv Kumar Math II

Definition: Let T:U→V be a linear map then Range of T R(T)= {v∈ V | v=Tu, for some u

∈U} The kernel(null space) of T is the set

ker(T) =N(T)={u∈U|T(u)=0v}.

n other words, N(T) is the set of all

those elements in U that are mapped Rajiv Kumar Math II

by T into the zero vector of V.

Theorem: Let T:U→V be a linear map. . R(T) is a subspace of V

i. N(T) is a subspace of U ii.T is one-one iff N(T) is the zero subspace of U. v If [{u1,u2,….,un}]=U, then R(T) =[{T(u1),T(u2),…T(un)}].

V. If U is finite-dimensional, then dim (R(T) )≤ dim U. Rajiv Kumar Math II

Proof: (i) Let v1, v2∈R(T). Then there exist vectors u1, u2 in U such that

T(u1)=v1 and T(u2)=v2. So 1+v2=T(u1)+T(u2)=T(u1+u2) as T is linear) But u1+u2∈U, since U is

vector space. Hence, v1+v2 is the image

n element of U. So v +v ∈R(T). Rajiv Kumar Math II

In the same way,αT(u1)=T(αu1), since T is linear. But αu1∈U, as U is a vector space. Hence, αv1∈ R(T).Thus R(T) is a subspace of V.

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(ii) To prove

N(T) is a subspace

of U Let u, v ∈N(T). Then T(u)= 0v and T(v)=0v, Now T(u + v)= T(u)+ T(v)=0v+0v =0v ⇒ u + v∈N(T)

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(iii) T is one-one iff N(T) is the zero subspace of U. Proof :

u ∈N(T) we want to prove u =0U . T(u)= 0v =T(0U) , as T is linear ⇒ u= 0U

as T is one-one .

⇒ N(T) is zero subspace of U Rajiv Kumar Math II

If N(T) =0U then T is one one T(u)= T(v) ⇒ T(u-v)= 0V ⇒u-v∈ N(T) ⇒ u-v= 0U (as N(T )=0U ) i.e. u=v ⇒ T is one-one.

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(iv) If [{u1,u2,….,un}]=U, then R(T) =[{T(u1),T(u2),…T(un)}]. We need to prove [{T(u1),T(u2),…T(un)}]⊆ R(T) ………..(I) & R(T)⊆[{T(u1),T(u2),…T(un)}]…… (II) Rajiv Kumar Math II

T(ui)∈R(T) ,∀i, 1≤i≤n ⇒[{T(u1),T(u2),…T(un)}]⊆ [R(T)] =R(T) As R(T) is the subspace

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v∈R(T) ⇒ ∃u∈U suct that v =Tu u∈U⇒ u =α1u1 +,... + α nun As

[{u1,u2,….,un}]=U,

T (u ) = α1T ( u1 ) + α 2T ( u2 ) +,...,+α nT ( un )

⇒R(T)⊆[{T(u1),T(u2),…T(un)}] ……(II) Rajiv Kumar Math II

v.If U is finite-dimensional, then dim (R(T) )≤ dim U. Let {u1,u2,…un } be a basis of U Then [{u1,u2,…un }]= U ⇒R(T) =[{T(u1),T(u2),…T(un)}] dim (R(T)) =dim [{T(u1),T(u2),…T(un)}] ≤ n =dim U. Rajiv Kumar Math II

Example

Let T: V4

→

P3 be a linear

map from defined as T ( x1, x2 , x3 , x4 ) = x1 + ( x2 − x3 ) x + ( x1 − x3 ) x

(i)

Find N(T) and a basis of N(T)

(ii) Find R(T) and a basis of R(T)

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3

{

N (T ) = ( x1, x2 , x3 , x4 ) x1 + ( x2 − x3 ) x + ( x1 − x3 ) x 3 = 0 P

3

}

( x1, x2 , x3 , x4 ) | x1 = 0, x2 − x3 = 0  =  & x1 − x3 = 0   =

{ (0,0,0, x4) | x4 is any real number }

= [(0,0,0,1)] Basis of N(T) is {(0,0,0,1)} Rajiv Kumar Math II

R (T ) = [{T (1,0,0,0), T (0,1,0,0), T (0,0,1,0), T (0,0,0,1)}] As {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} is a basis of V4.

T(1,0,0,0) = 1+x3 T(0,1,0,0) = 0+ (1-0)x =x T(0,0,1,0) = 0+ (0-1)x +(0-1)x3 T(0,0,0,1) = 0V

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Hence R(T) = [{1+x3 ,x, -x-x3, 0V }]

Hence R(T) = [{1+x3,x, -xx3}] Set {1+x3 ,x, -x-x3} is independe In P3, (prove that ) Hence {1+x3 ,x, -x-x3} is a basis of R(T) Rajiv Kumar Math II

Rank of a linear transformation T T : U→V, U finite dimensional Definition : dim R(T) is called rank(T) and is written as r(T).

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Nullity of a linear transformation T T : U→V, U finite dimensional Definition : dim N(T) is called nullity(T) and is written as n(T).

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Rank Nullity Theorem T : U → V , a linear transformation & U finite dimensional Then r(T) +n(T)= dim(U)

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Example :

Let T: V3

→

V4 be a linear map,

defined by T ( x1, x2 , x3 ) =

( x1, x2 − x3 , x1 − x2 + x3 , x1 + x2 − x3 )

Verify rank nullity theorem. .

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.

N (T ) = {( x1, x2 , x3 ) : T ( x1, x2 , x3 ) = (0,0,0,0)} ⇒ ( x1, x2 − x3 , x1 − x2 + x3 , x1 + x2 − x 3 ) = (0,0,0,0) ⇒ x1 = 0, x2 − x3 = 0, x1 − x2 + x3 = 0 x1 + x2 − x 3 = 0, Rajiv Kumar Math II

Solving system of four equations in three unknowns we get x1 =0 & x3 = x2 .

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N(T) ={(0,x3, x3) : x3 is arbitrary} =[{(0,1,1)}] R(T)= { (x1,x2 - x3,x1 -x2 +x3, x1 +x2 -x3):(x1,x2, x3)∈ V3}

= [{(1,0,1,1), (0,1,-1,1),(0,-1,1,-1)}] What is dim (R(T)) ? Rajiv Kumar Math II

R(T)= [{(1,0,1,1), (0,1,-1,1),(0,-1,1,1)}]

To find dimension of R(T) we find Linearly independent subset A of set S= {(1,0,1,1), (0,1,-1,1),(0,-1,1,1)} Such that [A]=[S]=R(T) Rajiv Kumar Math II

Theorem : If U & V are finite dimensional vector spaces of the same dimension , then a linear map T:U→ V is one-one iff it is onto.

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Question : If T : V4→ V3 is the linear transformation defined as T(1,0,0,0)= (-1,0,1) , T(0,1,0,0)= (3,-1,0) T(0,0,1,0)= (2,-1,1) T(0,0,0,1)= (1,0,-1) find ker(T) & r(T) ? Rajiv Kumar Math II

To find ker (T) we must find transformation T defined for any element of V4. T(x1,x2, x3 ,x4)= (-x1+3x2 +2x3+ x4, -x2 -x3, x1 +x3 -x4 )

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T(x1,x2, x3 ,x4)=(-x1+3x2 +2x3+ x4, -x2 -x3, x1 +x3 -x4

)

ker(T ) = {( x1, x2 , x3 , x4 ) : T ( x1, x2 , x3 , x4 ) = (0,0,0)} -x1+3x2 +2x3+ x4 =0 -x2 -x3, =0

x1 +x3 -x4 =0

Solving system of three equations in four unknowns we get x1 = -x3 +x4 , x2 =-x3` Rajiv Kumar Math II

Question : If T : V4→ V3 is the linear transformation defined as T(1,0,0,0)= (-1,0,1) , T(0,1,0,0)= (3,-1,0) T(0,0,1,0)= (2,-1,1) T(0,0,0,1)= (1,0,-1) find R(T) & n(T) ? Rajiv Kumar Math II