In (a) and (b) below, let S be the collection of vectors [x,y,z] in R3 that sati
ID: 2969983 • Letter: I
Question
In (a) and (b) below, let S be the collection of vectors [x,y,z] in R3 that satisfy the given property. Either prove that S forms a subspace of R3 or give a counterexample to show that it does not. Yoy think of the vectors as column vectors.
(a) x=y=z
(b) z=2x, y=0
In (a) and (b) below, let S be the collection of vectors [ x y z ] in R3 that satisfy the given property. Either prove that S forms a subspace of R3 or give a counterexample to show that it does not. You may think of the vectors as column vectors. x = y = z Z = 2x, y = 0Explanation / Answer
(a)
(i)
(0 0 0) is in S (hence S is nonempty)
(ii)
S is closed under addition, since (x y z), (p q r) are in S => x = y = z, p = q = r => x+p = y+q = z+r => (x+p y+q z+r) is in S
(ii)
S is closed under scalar multiplication since (x y z) is in S, k real => x = y =z => kx = ky = kz => (kx ky kz) is in S.
Thus S is a subspace.
(b)
(i)
(0 0 0) is in S (hence S is nonempty)
(ii)
S is closed under addition, since (x y z), (p q r) are in S => z = 2x,y=0, r = 2p, q = 0 => z+r = 2x+2p = 2(r+p), y+q = 0+0 = 0 => (x+p y+q z+r) is in S
(ii)
S is closed under scalar multiplication since (x y z) is in S, k real => z=2x, y = 0 => kz = k(2x) = 2(kx), ky = k.0 = 0 => (kx ky kz) is in S.
Thus S is a subspace.