Let V be the vector space of ordered pairs of complex numbers over the real field [RR] . Show
that V is of dimension 4.
Solution
Let S={(1,0),(0,1),(i,0),(0,i)} be the set . We wish to prove S is basis of V over the real field
R.Consequently we can prove that dim(V)=4, because number of elements in set S is 4.
Vectors in S are linearly independent. If
a(1,0)+b(0,1)+c(i,0)+d(0,i)=0
a+ic=0 => a=c=0
b+id=0 => b=d=0
a=b=c=d=0 only solution.
Thus vectors in S are linearly independent.
Let (x+iy,w+iz) in V
then
(x+iy,w+iz)=x(1,0)+y(i,0)+w(0,1)+z(0,i)
(x+iy,w+iz) is an arbitrary element of V. Thus every element of V can be expressed as linear
combination of vectors in S.
Thus S is basis for V. Number of elements in S are 4 ,so dimension of V be 4.
Hence proved..
Let V be the vector space of ordered pairs of complex numbers over t.pdf
1. Let V be the vector space of ordered pairs of complex numbers over the real field [RR] . Show
that V is of dimension 4.
Solution
Let S={(1,0),(0,1),(i,0),(0,i)} be the set . We wish to prove S is basis of V over the real field
R.Consequently we can prove that dim(V)=4, because number of elements in set S is 4.
Vectors in S are linearly independent. If
a(1,0)+b(0,1)+c(i,0)+d(0,i)=0
a+ic=0 => a=c=0
b+id=0 => b=d=0
a=b=c=d=0 only solution.
Thus vectors in S are linearly independent.
Let (x+iy,w+iz) in V
then
(x+iy,w+iz)=x(1,0)+y(i,0)+w(0,1)+z(0,i)
(x+iy,w+iz) is an arbitrary element of V. Thus every element of V can be expressed as linear
combination of vectors in S.
Thus S is basis for V. Number of elements in S are 4 ,so dimension of V be 4.
Hence proved.
![Let V be the vector space of ordered pairs of complex numbers over the real field [RR] . Show
that V is of dimension 4.
Solution
Let S={(1,0),(0,1),(i,0),(0,i)} be the set . We wish to prove S is basis of V over the real field
R.Consequently we can prove that dim(V)=4, because number of elements in set S is 4.
Vectors in S are linearly independent. If
a(1,0)+b(0,1)+c(i,0)+d(0,i)=0
a+ic=0 => a=c=0
b+id=0 => b=d=0
a=b=c=d=0 only solution.
Thus vectors in S are linearly independent.
Let (x+iy,w+iz) in V
then
(x+iy,w+iz)=x(1,0)+y(i,0)+w(0,1)+z(0,i)
(x+iy,w+iz) is an arbitrary element of V. Thus every element of V can be expressed as linear
combination of vectors in S.
Thus S is basis for V. Number of elements in S are 4 ,so dimension of V be 4.
Hence proved.](https://image.slidesharecdn.com/letvbethevectorspaceoforderedpairsofcomplexnumbersovert-230324024353-9fe071e0/85/let-v-be-the-vector-space-of-ordered-pairs-of-complex-numbers-over-tpdf-1-320.jpg)
