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### LINEAR TRANSFORMATION-THEORY & APPLICATIONS-IIT JAM PAPER SOLVED

#### THEORY & PROBLEMS ON LINEAR TRANSFORMATION

This is useful for those who are study B.Sc in mathematics & statistics & too preparing for IIT JAM,ISI M.MATH/M.STAT,NET,GATE & other exams.
Linear Mapping & Linear Transformation:
DEFINITION : Let V and W be vector spaces over the same field F. A mapping T: V->W is said to be a linear mapping if it satisfies the following conditions-
I. T(a+b)=T(a)+T(b) for all a,b€V.
II. T(ca)=cT(a)  for all c in F & for all a€V.
These two conditions can also be replaced by the single condition-
T(ma+nb)= mT(a)+nT(b) for all a,b€V & m,n in F.

NOTE : I. A linear mapping T:V->W is also a homomorphism of V to W.
II. Generally a liner mapping T is said is a transformation from one vector V to another vector space W, both over the same field of scalars.

EXAMPLES : 1. The Identity Mapping : The mapping T : V->V defined by T(a)=a for all a in V,is a linear mapping.
2. Let P be the vector space of all real polynomials. The mapping D:P->P defined by DP(x)=d/dx(P(x)) , P(x)€ P is a linear mapping.
3. Example of linear mapping: Let T:R^3->R^3 be defined by T(x,y,z)=(x,y,0), where (x,y,0)€R^3.
Sol : a=(x1,y1,z1), b=(x2,y2,z2)€ R^3.
Then a+b=(x1+x2,y1+y2,z1+z2)
Now T(a+b)=(x1+x2,y1+y2,0)=(x1,y1,0)+(y1,y2,0)=T(a)+T(b)
For c€R ca=(cx1,cy1,cz1)=c(x1,y1,z1)=cT(a).
Therefore T is a linear mapping.
4. Example of non-linear mapping: Let T:R^3->R^3 be defined by T(x,y,z)=(x+1,y+1,z+1) ,where (x,y,z)€R^3
Sol : a=(x1,y1,z1), b=(x2,y2,z2)€ R^3.
Then a+b=(x1+x2,y1+y2,z1+z2)
Now T(a+b)=(x1+x2+1,y1+y2+1,z1+z2+1)=(x1,y1,z1)+(y1,y2,z2)+(1,1,1)=/= T(a)+T(b)
Therefore T is not a linear mapping.

Properties of Linear Transformation :
Theorem : Let V and W be vector spaces over the same field F &  T: V->W be a linear mapping.Then
(I) T(-a)=-T(a) for all a€V.
(II) T(a-b)=T(a)-T(b) for all a,b€V.
(III) T(k1a1+k2a2+k3a3+ . . . . . +knan)=k1T(a1)+k2T(a2)+ . . . . +knT(an); where ai€V & ki€ F ,i=1(1)n.
Proof : (I) T(a+b)=T(a)+T(b)
Put b= -a then the proof is complete.
(II) T(a-b)=T(a)+T(-b)=T(a)-T(b)
(III) It can be done by induction.

Kernel of T or Null Space of T :-
Let V and W be vector spaces over a field F. Let T : V-> W be a linear mapping. The set of all vectors a€V such that T(a)=0, 0 being the null vector in W,is said to be the Kernel of T or the Null Space of T,and denoted by Ker(T) or N(T).
Ker(T)={ a€V : T(a)=0},where 0 being the null vector.

Theorem: T: V->W is a linear mapping,then T is injective or one-one mapping if & only if Ker(T)={0},where 0 being the null vector.
Proof : Only if part:- Let T be injective,then T(0)=0 is the only pre image,where 0 being the null vector. Ker(T)={0}.
If part :- Ker(T)={0} ,and a,b be two elements in V s.t. T(a)=T(b) in W.
Implying T(a)-T(b)=T(a-b)=0,since T is linear.
That is, a-b€Ker(T) and since Ker(T)={0} ,So a=b.
Thus T(a)=T(b) implying a=b & therefore T is injective.

Images of a linear mapping :-
Let V and W be vector spaces over a field F. Let T:V->W be a linear mapping. The images of the elements of V under the mapping T form a subset of W. This subset is said to be the image of T & is denoted by ImT or it is also called Range of T (R(T)).
Im(T) or R(T)={T(a) : a€V }.
Theorem : Let V and W be vector spaces over a field F. Let T:V->W be a linear mapping.Then ImT  is a subspace of W & KerT is a subspace of V.
Proof : Try Urself.

Theorem : Let T:V->W be a linear transformation. If {a1,a2, . . . ,an} is a basis for V & T is one-to-one then {T(a1),T(a2), . . . . ,T(an)} is a basis for W.
Proof :- To show {T(a1),T(a2), . . . . ,T(an)} is a basis for W we need to check the linear independence & generating property of of the vectors T(a1),T(a2), . . . . ,T(an).
Consider the relation
c1T(a1)+c2T(a2)+ . . . . +cnT(an)=0 ( 0 denotes zero vector),where ci€F for all i=1(1)n.
=> T(c1a1+c2a2+ . . . . +cnan)=0,since T is linear.
=> c1a1+c2a2+ . . . . +cnan=0
=> c1=c2= . . . =cn=0,since {a1,a2, . . . ,an} is a basis for V.Therefore
{T(a1),T(a2), . . . . ,T(an)} is linearly independent.
Check that {T(a1),T(a2), . . . . ,T(an)} is also a spanning set in W. So, they form a basis for W.

Remark : If T:V->W be a linear mapping such that dim(V)=dim(W) then the followings are equivalent:
A. T is non-singular;    B.T is invertible;   C.T is one-to-one;    D.T is onto;

Solved Examples:-

Ex.1. Determine T:R^3->R^3 which maps the basis vectors e1,e2,e3 of R^3 to (1,1),(2,3),(-1,2) of R^2. Find T(x).
Sol : x=x1e1+x2e2+x3e3, here x is a vector
T(x)=x1T(e1)+x2T(e2)+x3T(e3)=x1(1,1)+x2(2,3)+x3(-1,2)=(x1+2x2-x3,x1+3x2+2x3).

Ex.2. Show that a mapping T: R^3->R^3 is defined by
T(x,y,z)=(x+y,y+z,z+x) is a linear transformation.Also find the dimension of Ker(T) & also the transformation is one-one. Find ImT.Is it onto?
Sol : I. Show T(ma+nb)=mT(a)+nT(b),where T(a)=(x1+y1,y1+z1,x1+z1) & T(b)=(x2+y2,y2+z2,x2+z2).
II. KerT={x : T(x)=0} ,here x is a vector.
So, T(x)=0 implying x+y=y+z=z+x=0, i.e. x=y=z=0. KerT={0} and hence T is one-one. Therefore dim(kerT)=0.
III. Let {e1,e2,e3} be a basis for R^3.
=> {T(e1),T(e2),T(e3)} is a basis for ImT.
=> {(1,0,1),(1,1,0),(0,1,1)} is a basis for ImT.
ImT=K*{(1,0,1),(1,1,0),(0,1,1)}€R^3, so T is an onto mapping.

Ex.3. (IIT JAM-2007) Let T:R^3->R^3 be a linear mapping defined by T(x,y,z)=(x+y,x-z).Then find the dimension of the null space of T is
A.0         B.1         C.2         D.3
Sol : (B) T(x,y,z)=(x+y,x-z).
Now (x+y,x-z)=(0,0)
=> x=-y=z
Then N(T)={a(1,-1,1)|a€R}
i.e. dimension of the null space of T is 1.

Nullity & Rank of a linear mapping:-
Let V and W be vector spaces over a field F and T:V->W be a linear mapping. Then ker(T) is a subspace of V.Then the dimension of Ker(T) is called the nullity of T. ImT is a subspace of W. Then the dimension of ImT is called the rank of T.
Rank(T)=dim(ImT), Nullity(T)=dim(kerT)

Theorem: Let V and W be the vector spaces over a field F and V is finite dimensional. If T:V->W be  linear mapping then Nullity of T+Rank of T=dimV
Proof: (not necessary)see any standard book.

Ex. Determine the linear mapping :R^3->R^3 which maps the basis vectors (0,1,1),(1,0,1),(1,1,0) of R^3 to (1,1,1),(1,1,1),(1,1,1) respectively.Verify that dim(kerT)+dim(ImT)=3.
Sol:- let ci be unique scalars,then
k(is a vector)=c1(0,1,1)+c2(1,0,1)+c3(1,1,0)
Then c2+c3=x, c1+c3=y,  c1+c2=z.
Solving we get c1=1/2*(y+z-x) , c2=1/2*(z+x-y) , c3=1/2*(x+y-z).
Since T is linear,
T(k)=c1T(0,1,1)+c2T(1,0,1)+c3(1,1,0)
= c1(1,1,1)+c2(1,1,1)+c3(1,1,1)
=(c1+c2+c3,c1+c2+c3,c1+c2+c3)
=> T(x,y,z)=(1/2*(x+y+z),1/2*(x+y+z),1/2*(x+y+z)); (x+y+z)€R^3.
To find kerT,put T(x,y,z)=(0,0,0)
x+y+z=0, let y=c,z=d,x=-(c+d).
(x,y,z)=(-c-d,c,d)=c(-1,1,0)+d(-1,0,1)
Ker(T)=L{(-1,1,0)+(-1,0,1)} since (-1,1,0),(-1,0,1) are linearly independent,so dim(kerT)=2.
Since (0,1,1),(1,0,1),(1,1,0) form a basis of R^3,
ImT=L{(1,1,1)}. Hence dim(ImT)=1.
So hence the proof is complete.

Composition of Linear mapping :-
Let V,W and U be vector spaces over a field F and let T:V->W & S:W->U be a linear mapping. The composite mapping SoT:V->U is defined by SoT(a)=S{T(a)},a€V.
It can easily be shown that the composite mapping is linear.

Inverse Mapping:-
A linear mapping T is invertible if & only if T is one-one & onto.
Let T:V->W is invertible then T^(-1) : W->V is linear(proof is done by using the definition of linear mapping).
A linear mapping is said to be non-singular if it is invertible.

Ex. A linear mapping T:R^3->R^3 is defined by T(x,y,z)=(x-y,x+2y,y+3z),(x,y,z)€R^3. Show that T is non-singular & determine T^(-1).
Sol:- T is a linear mapping.
T(x,y,z)=(0,0,0) => x=y=z=0.
Ker(T)={0}. T is one-one.
And dim(V)=dim(W)=3, so T is onto.
So,it is non-singular & invertible.
Let T^(-1)(x,y,z)=(a,b,c)
=> T(a,b,c)=(x,y,z)=(a-b,a+2b,b+3c)
Then a-b=x , a+2b=y , b+3c=z
Solve for a,b,c & you will get the answer.

Matrix representation of a linear mapping:-
Ex.1. A linear T:R^3->R^2 is defined by   T(x,y,z)=(3x-2y+z,x-3y-2z) ,(x,y,z)€R^3.
Find the matrix of T relative to the ordered bases
{(1,0,0),(0,1,0),(0,0,1)} of R^3 and {(1,0),(0,1)} of R^2.Also find rank(T).
Sol : T(1,0,0)=(3,1)=3(1,0)+1(0,1)
T(0,1,0)=(-2,-3)=-2(1,0)-3(0,1)
T(0,0,1)=(1,-2)=1(1,0)-2(0,1)
So the matrix of T is   ( 3    -2    1 )
( 1    -3   -2 )
The rank of these 2*3 matrix is 2,so rank(T)=2.

Ex.2. (IIT JAM-2010) Consider the transformation T: R^3 -> R^3 given by
T(x,y,z)=(kx+y+z,x+ky+z,x+y+kz).
Find the matrix P associated with the above transformation with respact to the standard basis. Further,find the values of k for which P has zero as one of its eigen values.
Sol : {(1,0,0),(0,1,0),(0,0,1)} be the standard basis.
Now T(1,0,0)=(k,1,1)=ke1+1e2+1e3
T(0,1,0)=(1,k,1)=1e1+ke2+1e3
T(0,0,1)=(1,1,k)=1e1+1e2+ke3
Hence the associated matrix P(3*3) is given by
P  = ( k  1  1 )
( 1  k  1 )
( 1  1  k )
Now put k=1,then see that P has zero as one of its eigen values.