Let A x y z w F4 x 2y z w 0 and x 2y z w 0 a Prove

Let A = {(x, y, z, w) F^4| x + 2y - z + w = 0 and x - 2y + z - w = 0} (a) Prove that A is a subspace of F^4. (b) Find a basis for A, and compute dim(A). (c) Extend your basis of A to a basis of F^4.

Solution

A is an ordered elements of four such that

they have the relation as

x+2y-z+w =0 and

x-2y+z-w=0

We have 4 variables and 2 independent equations.

Solutions will be of 2 free variables and 2 dependent variables.

Adding these two, we get 2x=0 or x=0

2y-z+w =0 is the relation between the 3 variables

Let y =y

z=z

w = z-2y

Hence solutions would be

all 4 ordered numbers of the form

(0, y, z, z-2y)

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Consider any two elements in A

say (0,y1, z1, z1-2y1) and (0,y2, z2, z2-2y2)

Adding we get (0,y1+y2, z1+z2, z1+z2-2(y1+y2)

Hence this again belongs to A.

Closure is true.

Identity is (0,0,0,0)

Inverse is (0, -y, -z, -z+2y) for any element (0,y,z,z+2y)

Hence A forms a subspace

b) Dimension of A would be

the free variables in the pattern hence 2.

In (0,y,z,2z-y) there are only 2 free variables.

Basis = (0 1 0 -1) and (0,0, 1, 2)

3) F^4 will be as (x,y,z,w) with no relation between x,y,z,w

Hence dim = 4

So base would be

(1,0,0,0) (0,1,0,0) (0,0,1,0) (0,0,0,1) for F4