Prove that given a vector subspace S of Rn suptac the set of

Prove that given a vector subspace S of R^n, s^uptac, the set of all vectors orthogonal to every vector of S, is also a vector subspace of R^n. Construct the spanning sets of the orthogonal subspace s^uptac to the subspaces S defined by the following spanning sets: (a) K = {(-2, 3)} (b) K = {(-2, 3, 3) (c) K = (-2, 3, 3), (0, 3, -7)

Solution

4. Let x, y S be any two arbitrary vectors and let c be an arbitrary scalar. Also, let z S be an arbitrary vector. Then (x+y).z = xz+yz = 0. Further, (cx).z = c(x.z)= c.0 = 0. Hence x+y and c x S ,so that S is a vector space and hence, a subspace of Rⁿ.

5. (a). S = span {(-2,3)}. Let (x,y) be an arbitrary vector in S. Then (x,y).(-2,3) = 0 or -2x+3y = 0 or, x = 3y/2. Hence (x,y) = (3y/2,y) = y(3/2,1). Therefore, S = span {(3/2,1)}.

      (b). S = span {(-2,3,3)}. Let (x,y,z) be an arbitrary vector in S. Then (x,y,z).(-2,3,3) =0 or, -2x+3y+3z=0 or, x = 3y/2+3z/2 . Therefore, (x,y,z) = (3y/2+3z/2, y,z) = y(3/2,1,0)+z(3/2,0,1). Hence S = span {(3/2,1,0), (3/2,0,1)}.

     (c ). S = span {(-2,3,3), (0,3,-7)}. Let (x,y,z) be an arbitrary vector in S. Then (x,y,z).(-2,3,3) =0 or, -2x+3y+3z=0. Also, 3y-7z = 0. Let z = 6t. Then y = 14t and x = 3y/2+3z/2= 21t+ 9t= 30t so that (x,y,z) = t( 30,14,6) = 2t(15, 7,3). Hence, Hence S = span {(15, 7,3)}.