# 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}.

(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.

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.

## Comments

## Leave a comment