Question

Let U = {(x1,x2,x3,x4) ∈F4 | 2x1 = x3, x1 + x4 = 0}.

(a) Prove that U is a subspace of F4.

(b) Find a basis for U and prove that dimU = 2.

(c) Complete the basis for U in (b) to a basis of F4.

(d) Find an explicit isomorphism T : U →F2.

(e) Let T as in part (d). Find a linear map S: F4 →F2 such that S(u) = T(u) for all u ∈ U.

Answer 1

We have U = {(x₁,x₂,x₃,x₄) ∈ F⁴ | 2x₁ = x₃, x₁ + x₄ = 0}. Then, (x₁,x₂,x₃,x₄) = (x₁,x₂,2x₁, -x₁).

(a).Let X=(x₁,x₂,2x₁,-x₁)and Y=(y₁,y₂,2y₁,-y₁) be 2 arbitrary vectors in U and let k be an arbitrary scalar ∈F. Then X+Y=(x₁,x₂,2x₁, -x₁)+(y₁,y₂,2y₁, -y₁) = (x₁+y₁, x₂+y₂, 2(x₁+y₁),-(x₁+y₁)). This implies that X+Y ∈ U so that U is closed under vector addition. Further, kX = k(x₁,x₂,2x₁, -x₁) = (kx₁,kx₂,2kx₁, -kx₁). This implies that kX ∈ U so that U is closed under scalar multiplication. Also, it is apparent that the zero vector (0,0,0,0) ∈ U. Hence U is a vector space. Since U ⊆ F⁴, hence U is a subspace of F⁴.

(b).An arbitrary vector in U is of the form (x₁,x₂,2x₁, -x₁) = x₁(1,0,2,-1)+x₂( 0,1,0,0). This implies that {(1,0,2,-1),( 0,1,0,0)} is a basis for U. Further, dim(U) = 2.

(c ). Let A =

1	0
0	1
2	0
-1	0

The RREF of A is

1	0
0	1
0	0
0	0

It implies that {(1,0,2,-1),( 0,1,0,0),(0,0,1,0),(0,0,0,1)} is a basis for F⁴.

Please post the parts (d) and (e) again separately.