# Let V be the vector space of ordered pairs of complex numbers over the real field `RR` . Show that V is of dimension 4.

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### 1 Answer

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.