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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 &ge; 0}; (iii) {(a,b,c) : a = &minus;b}.
1
Expert's answer
2012-10-24T09:46:48-0400
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&#039;b&#039;c&#039;) belong
to L and t belongs to R,
so
a+b+c = 0,
a&#039;+b&#039;+c&#039; = 0.
Then

x+y=(a+a&#039;, b+b&#039;, c+c&#039;),
and
a+a&#039; + b+b&#039; + c+c&#039; = a+b+c +
a&#039;+b&#039;+c&#039; = 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 &ge;
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 =
&minus;b}. Then L is subspace of R3.
Indeed, let
x=(a,b,c), and y=(a&#039;b&#039;c&#039;)
belong to L and t belongs to R,
so
a=-b ,
a&#039;=-b&#039;.
Then

x+y=(a+a&#039;, b+b&#039;, c+c&#039;),
and
a+a&#039; = -b-b&#039; = -(b+b&#039;)

Also

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

Thus L is a subspace of
R3.

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