Homework SourseConsider R4 in the following set W x1 x2 x3 x4 elementof R4
Consider R^4 in the following set:. W = {(x_1, x_2, x_3, x_4) elementof R^4: 3x2 - 2x_4 = 0} Show w that is subspace a of R^4
Solution
We have W = {(x1,x2,x3,x4) R4: 3x2-2x4 =0}. Let X = (x1,x2,x3,x4) and Y = (y1,y2,y3,y4) be two arbitrary elements in W and let c be an arbitrary scalar. Then (X+Y) = (x1+y1,x2+y2,x3+y3,x4+y4). Also, 3(x2+y2) - 2 (x4+y4) = (3x2-2x4) + (3y2-2y4) = 0+0= 0 as 3x2-2x4= 0 and 3y2-2y4 = 0. Hence W is closed under vector addition. Further, cX = (cx1,cx2,cx3,cx4) and 3cx2-2cx4 = c(3x2-2x4) = c*0 = 0. Hence W is closed under scalar multiplication. Further, the zero vector = (0,0,0,0) obviously belongs to W as 3*0-2*0 = 0. Hence W is a vector space and therefore, W is a subspace of R4.
