Why isn't the set {(0,0,0),(1,1,1)} a subspace in R3?

Which one of the following sets of vectors is a subspace of R3?


A: {(0; 0; 0); (1; 1; 1)}


B: {(a; a + 1; a) | a is a real number}


C: {(a; b; c) | a^2 + b^2 + c^2 = 16}


D: the set of all vectors in R3 except the vector (1; 0; 1)


E: {(a; a + b; b) | a; b are real numbers}





the answer is E, but I can't figure out why!?


thanks for help in advance

Why isn't the set {(0,0,0),(1,1,1)} a subspace in R3?
A isn't because it isn't closed under scalar multiplication (2(1,1,1) = (2,2,2) which is not a member of the set. The set only contains two points, (0,0,0) and (1,1,1).





B isn't because it isn't closed under addition (1,2,1) and (2,3,2) are in the set, but (3,5,3) isn't. (3,5,3) is not a member because it is not of the form (a, a+1,a)





C isn't because it is not closed under addition (4,0,0) and (0,0,4) are members but (4,0,4) isn't (4,0,4) is not a member because 4^2+0^2+4^2 = 32 and not 16





D isn't because it is not closed under addition (1,1,1) and (0,1,0) are in the set, but (1,1,1) + -(0,1,0) isn't





For question 44, the reason the answer is B is you know that -2(1,1,1) = (-2,-2,-2) and -1(1,1,1) = (-1,-1,-1) are both in the set. However, when you add these two together, you get (-3,-3,-3) = -3(1,1,1). This is not in your set, because the smallest that a can be is -2. Hence, this space is not closed under addition, and thus can not be a subspace

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