Using the closure properties of subspaces, determine whether the set is a subspa

Question

Using the closure properties of subspaces, determine whether the set is a subspace of R^3. If not, indicate which properties fail.
a. All vectors of the form (a,0,0)
b. All vectors with integer components
c. All vectors (a,b,c) for which b = a + c
d. All vectors (a,b,c) for which a + b + c = 1
e. All vectors with exactly one non-zero component
f. All x which for the norm of x <= 1
g. All vectors of the form (a,-a,c)
h. All vectors (a,b,c) for which a + b + c = 0

And can you give a reason/method as to how you got your answer, that would be great! :D

Explanation / Answer

Here are the basic things you need to check: 1)is there a zero vector 2)is it closed under addition 3)is it closed under multiplication So here we go: a) 1) if a=0, then (0,0,0) 2) (a1,0,0) + (a2,0,0) = (a1+a2,0,0) = (a3,0,0) which has the right form 3) (a,0,0)*r = (a*r,0,0) = (a1,0,0) which has the right form. answer Yes b) All vectors with integer components ~ (x,y,z) where x,y,z are integers 1)if x,y,z = 0, then (0,0,0) exists 2)(x1,y1,z1) + (x2,y2,z2) = (x1+x2, y1+y2, z1+z2) = (x3,y3,z3) which has the right form 3)(x,y,z)*r = (x*r, y*r, z*r) = (x1, y1, z1) which has the right form answer Yes (I'll explain if it is wrong now) c) (a,b,c), where b= a+c, so (a, a+c, c) answer Yes d) (a,b,c) where a+b+c = 1 if a=0 and b=0, then c=0 isn't true, there is no zero vector answer no e)(a,0,0) or (0,b,0) or (0,0,c) Same as a answer yes f) (second opinion on this question might help, i'm not 100% sure) Answer yes g)(a, -a, c) Answer yes h) (a,b,c) where a+b+c=0 Answer Yes

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