Homework SourseQ2 Is x y z w R4 3x 2y z 0 and x z 3w 0 a vector subspa
Q2) Is {(x, y, z, w) R4 : 3x + 2y + z = 0 and x z 3w = 0} a vector subspace of R 4 ? If so, find a basis of it.
Solution
Let the set {(x, y, z, w)T R4: 3x+2y+z =0 and xz3w= 0} be denoted by V. Further, let (x1,y1,z1,w1)T and (x2,y2,z2,w2)T be two arbitrary elements of V and let c be an arbitrary scalar. Then 3x1 + 2y1 + z1 = 0…(1), x1 – z1 3w1 = 0…(2), 3x2 + 2y2 + z2 = 0…(3) and x2 – z2 3w2 = 0…(4).
Now, on adding the 1st and the 3rd equations, we get 3x1 + 2y1 + z1+3x2 + 2y2 + z2 = 0 or, 3(x1+x2) +2(y1+y2) +(z1+z2) = 0…(5). Similarly, on adding the 2nd and the 4th equations, we get x1 – z1 3w1+ x2 – z2 3w2 = 0 or, (x1+x2) –(z1+z2) -3(w1+w2) = 0...(6). This means that (x1,y1,z1,w1)T+(x2,y2,z2,w2)T = ( x1+x2, y1+y, z1+z2, w1+w2)T V. Thus, V is closed under vector addition.
Further, on multiplying both the sides of the 1st and the 2nd equations by c, we get 3cx1 + 2cy1 + cz1 = 0..(7) and cx1 –c z1 – 3cw1 = 0…(8) so that c(x1,y1,z1,w1)T = (cx1,cy1,cz1,cw1)T V. Thus, V is closed under scalar multiplication. Hence V is a vector space.
Let z = r and w = 2t. Then x = r +6t and y = ½[ -3(r +6t) – r] = ½( -4r -18t) = -2r-9t. Then (x,y,z,w)T = (r +6t, -2r-9t, r,t)T = r( 1,-2,1,0)T + t( 6,-9,0,1)T. Thus, V = span{( 1,-2,1,0)T, ( 6,-9,0,1)T }. Therefore, {( 1,-2,1,0)T, ( 6,-9,0,1)T } is a basis for V.
