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Answer to Question #16949 in Algebra for Ciaran Duffy

Question #16949
Decide (giving your reasons) whether the following sets are subspaces of R3:
(i) {(a,b,c) : a+b+c = 0}; (ii) {(a,b,c) : a ≥ 0};
(iii) {(a,b,c) : a = −b}.
Expert's answer
A subset L of R3 is a subspace if for any
x,y from L and real t
we
have that
x+y, t*x belong to L

(i) Let L = {(a,b,c) : a+b+c = 0}.
Then L is subspace of R3.
Indeed, let
x=(a,b,c), and y=(a'b'c') belong
to L and t belongs to R,
so
a+b+c = 0,
a'+b'+c' = 0.
Then

x+y=(a+a', b+b', c+c'),
and
a+a' + b+b' + c+c' = a+b+c +
a'+b'+c' = 0

Also
tx = (ta,tb,tc)
and
ta+tb+tc=t(a+b+c) =
t*0=0

Thus L is a subspace of R3.


(ii) Let L={(a,b,c) : a ≥
0}.
Then L is not a subspace of R3, since x=(1,0,0) belong to L, but for
t=-1
tx=(-1,0,0) does not belong to L


(iii) Let {(a,b,c) : a =
−b}. Then L is subspace of R3.
Indeed, let
x=(a,b,c), and y=(a'b'c')
belong to L and t belongs to R,
so
a=-b ,
a'=-b'.
Then

x+y=(a+a', b+b', c+c'),
and
a+a' = -b-b' = -(b+b')

Also

tx = (ta,tb,tc)
and
ta = t(-b) = -tb

Thus L is a subspace of
R3.

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