Question:-
Show that the mapping T:V2(R)→V3(R)
defined by T(a,b)=(a+b, a-b, b) is a linear transformation from V2(R)
into V3(R).
Find the range, rank, null space and
null(T) of T.
Solution:-
Given that
T:V2(R)→V3(R)
Such that
T(a,b)=(a+b,a-b,b)
Then aα+bβ= a(a1,b1)+b(a2,b2)
=(aa1,ab1)+(ba2,bb2)
=(aa1+ba2, ab1+bb2)
∴
T(aα+bβ)= T(aa+ba, ab+bb)
= (aa1+ba2+ab1+bb2,
aa1+ba2-ab1-bb2, ab1+bb2)
=[a(a1+b1)+b(a2+b2),
a(a1-b1)+b(a2-b2), ab1+bb2]
=a(a1+b1, a1-b1,
b1)+b(a2+b2, a2-b2, b2)
=aT(α)+bT(β)
∴ T:V2(R)→V3(R) is a linear
transformation.
Since T:V2(R)→V3(R)
∴
The set {(1,0), (0,1)} is a standard basis of V2(R).
∴ T(1,0) = (1, 1, 0)
T(0,1) = (1, -1, 1)
The set
{(1, 1, 0), (1, -1, 1)} span range of T.
Also if
a, b∊R such that
a(1, 1, 0)+b(1, -1, 1)=0
⇒ (a, a, 0)+(b, -b, b)=0
⇒ (a+b, a-b, b)=(0, 0, 0)
or a+b=0
& a-b=0
& b=0
⇒ a=0, b=0
∴
The set {(1, 1, 0), (1, -1, 1)} is Linear transformation.
∴
The set {(1, 1, 0), (1,-1, 1)} will be a basis of range of T.
∴ dimR(T)=2
∴ Rank(T)=2
∵ Rank(T)+Nullity(T)=dimV2(R)
⇒ 2+nullity(T)=2
⇒ nullity(T)=0
∴
Null space of T is only zero subspace of V2(R).
0 comments:
Post a Comment