Let x x1 x2 x3 x4 be a vector in R4 There are 24 ways to re

Let x = (x_1, x_2, x_3, x_4) be a vector in R^4. There are 24 ways to rearrange these 4 entries into another vector, like (x_2, x_1, x_3, x_4) and (x_4, x_3, x_2, x_1). Those 24 vectors, including z itself, span a subspace S. Pick specific vectors x so that the dimension of that subspace is (a) zero, (b) one, (c) three, (d) four.

Solution

a . when x= (0,0,0,0) then the subspace will have only \'x \' asits element and its dimension is 0

b . when x= ( a,0,0,0)

then the subspace S consists of all elements of the form = { (a,0,0,0) / a isin R }

Basis of S is B₁ = { ( 1,0,0,0) } and the dimension is 1 .

c =. when x = ( 1,2,3,0)

Basis of the subspace B₂ = { (1,0,0,0) , (0,2,0,0) , (0,0,3 ,0) , (0,0,0,0) }   

hence the dimension =3

d when X= (1,2,3,4)

basis of the subapce = B₃= { ( 1,0,0,0) , (0,2,0,0) ,(0,0,3,0) , (0,0,0,4) }

dimension of the subspace =4