Question 836555
Remember, a set is a subspace of *[tex \mathbb{R}^n] if it is closed under addition and scalar multiplication.


a) is not a subspace because (x1, 1, -6) + (x2, 1, -6) = (x1+x2, 2, -12) which is not in the original set.

b) is not a subspace since we can scalar multiply an element (x,y,z) by a large enough constant so that -7x^2 - 4y^2 - 8z^2 > 1.


c) is a subspace since (0, 6y_1, 0) + (0, 6y_2, 0) = (0, 6(y_1 + y_2), 0), and k(0, 6y, 0) = (0, 6ky, 0).


d) is a subspace since the condition implies y = z; you can check that the closure conditions hold.