Determine whether or not the following are subsets or subspa

Determine whether or not the following are subsets or subspaces of the given vector space V:

1) (a) S = {(x, x+1) | x} V = ²

(b) S = { (x₁, x₂, x₃) | x₁₁x₂ x₃= 3x₁+2} V = ²

(c) S = {all permutation matrices P} V = set of all square matrices

Solution

Solution: Observe that for (x, y) R 2 , [(x, y)]B1 = x y and [(x, y)]B2 = x+y 2 xy 2 . Also, T(1, 0) = (1, 1), T(0, 1) = (1, 2), T(1, 1) = (2, 1) and T(1, 1) = (0, 3). Thus, we have T[B1, B1] = [T 1, 0) ]B1 , [T 0, 1) ]B1 = [(1, 1)]B1 , [(1, 2)]B1 = 1 1 1 2 and T[B2, B2] = [T 1, 1) ]B2 , [T 1, 1) ]B2 = [(2, 1)]B2 , [(0, 3)]B2 = 1 2 3 2 3 2 3 2 . Hence, we see that [T(x, y)]B1 = (x + y, x 2y) B1 = x + y x 2y = 1 1 1 2 x y and [T(x, y)]B2 = (x + y, x 2y) B2 = 2xy 2 3y 2 = 1 2 3 2 3 2 3 2 x+y 2 xy 2 3. Let B1 = (1, 0, 0),(0, 1, 0),(0, 0, 1) and B2 = (1, 0),(0, 1) be ordered bases of R 3 and R 2 , respectively. Define T : R 3R 2 by T(x, y, z) = (x + y z, x + z). Then T[B1, B2] = [(1, 0, 0)]B2 , [(0, 1, 0)]B2 , [(0, 0, 1)]B2 = 1 1 1 1 0 1 . Check that [T(x, y, z)]B2 = (x + y z, x + z) t = T[B1, B2] [(x, y, z)]B1 . 4. Let B1 = (1, 0, 0),(0, 1, 0),(0, 0, 1) , B2 = (1, 0, 0),(1, 1, 0),(1, 1, 1) be two ordered bases of R 3 . Define T : R 3R 3 by T(x t ) = x for all x t R 3 . Then [T(1, 0, 0)]B2 = 1 · (1, 0, 0) + 0 · (1, 1, 0) + 0 · (1, 1, 1) = (1, 0, 0)t , [T(0, 1, 0)]B2 = 1 · (1, 0, 0) + 1 · (1, 1, 0) + 0 · (1, 1, 1) = (1, 1, 0)t , and [T(0, 0, 1)]B2 = 0 · (1, 0, 0) + (1) · (1, 1, 0) + 1 · (1, 1, 1) = (0, 1, 1)t . Thus, check that T[B1, B2] = [[T(1, 0, 0)]B2 , [T(0, 1, 0)]B2 , [T(0, 0, 1)]B2 ] = [(1, 0, 0)t , (1, 1, 0)t , (0, 1, 1)t ] = 1 1 0 0 1 1 0 0 1 , T[B2, B1] = [[T(1, 0, 0)]B1 , [T(1, 1, 0)]B1 , [T(1, 1, 1)]B1 ] = 1 1 1 0 1 1 0 0 1 , T[B1, B1] = I3 = T[B2, B2] and T[B2, B1] 1 = T[B1, B2]. 4