Vector Spaces: Determining Subspaces in R^3 and R^2

[mccoy1]

Homework Statement

Which of the following are a vector spaces?
(a)R = {(a,b,c) 〖 ∈R〗^3 │ a+b=0 and 2a-b-c=0}
(b)S = {(a,b)∈ R^2│ a^2=b^2 }..., can I say a =b which will make things simpler?
(c)T = {(f∈P_2 (R)│ f(1)=f(0)+1}

Homework Equations

The Attempt at a Solution

(a)
(0) 0(a,b,c)=(0,0,0)∈S because a+b=0+0=0 and
2a-b-c=0=2(0)-0-0=0
(A)Closure under addition:Take any (a_1,b_1,c_1 ) and (a_2,b_2,c_2 ) in R.
(a_1+b_1 )+(a_2+b_2 )=(a_1+a_2 )+(b_1+b_2 )=0
and (2a_1,-b_1,〖-c〗_1 )+(〖2a〗_2,〖-b〗_2,- c_2 )
=(2(a_1+a_2 ),〖-(b〗_1+b_2),-(c_1-c_2 ))
=2c-d-e=0|c,d,e∈R and c=(a_1+a_2),d=(b_1+b_2),e=-(c_1+c_2)
Which is in R
(B) Let k be a constant (real) and (a3,b3,c3) be a vector in S, So k(a3,b3,c3) is also in R. Also ka3+kb3 =0 and 2ka3-kb3-kc3 = 0.

(b)(0) If we take a=0 and b=0,
then (a,b)=(0,0)=z is an element of S ,so S ≠⊘ and property (0) is satisfied.
(A) Take any v_1=(a_1,b_1 ) and v_2= (a_2,b_2 )to be two vectors in S
v_1+v_2=(a_1,b_1 )+(a_2,b_2 )=(a_1+a_2,b_1+b_2 ).
(a_1+a_2 )^2=(b_1+b_2 )^2=(a_1^2+〖2a_1 a〗_1+a_2^2 )=(b_1^2+〖2b_1 b〗_2+b_2^2 )
c=a_1+a_2,d= b_1+b_2,for c,d ∈R.So c^2=d^2.This is not the same as a^2 =b^2, so S isn't a subspace of R^2 ...?
(B)
(c) not a subspace..
In conclusion, I have only one subspace, which is part (a)...? Also, in (a)confusion: a+b =0 is R^2, not R^3...
I'd appreciate your great help as always. I'm studying this by myself and so no teacher I can ask for help.
Cheers.

 

[lanedance]

Homework Statement

Which of the following are a vector spaces?
(a)R = {(a,b,c) 〖 ∈R〗^3 │ a+b=0 and 2a-b-c=0}
(b)S = {(a,b)∈ R^2│ a^2=b^2 }..., can I say a =b which will make things simpler?
(c)T = {(f∈P_2 (R)│ f(1)=f(0)+1}

Homework Equations

The Attempt at a Solution

(a)
(0) 0(a,b,c)=(0,0,0)∈S because a+b=0+0=0 and
2a-b-c=0=2(0)-0-0=0
(A)Closure under addition:Take any (a_1,b_1,c_1 ) and (a_2,b_2,c_2 ) in R.
(a_1+b_1 )+(a_2+b_2 )=(a_1+a_2 )+(b_1+b_2 )=0
and (2a_1,-b_1,〖-c〗_1 )+(〖2a〗_2,〖-b〗_2,- c_2 )
=(2(a_1+a_2 ),〖-(b〗_1+b_2),-(c_1-c_2 ))
=2c-d-e=0|c,d,e∈R and c=(a_1+a_2),d=(b_1+b_2),e=-(c_1+c_2)
Which is in R
(B) Let k be a constant (real) and (a3,b3,c3) be a vector in S, So k(a3,b3,c3) is also in R. Also ka3+kb3 =0 and 2ka3-kb3-kc3 = 0.

(b)(0) If we take a=0 and b=0,
then (a,b)=(0,0)=z is an element of S ,so S ≠⊘ and property (0) is satisfied.
(A) Take any v_1=(a_1,b_1 ) and v_2= (a_2,b_2 )to be two vectors in S
v_1+v_2=(a_1,b_1 )+(a_2,b_2 )=(a_1+a_2,b_1+b_2 ).
(a_1+a_2 )^2=(b_1+b_2 )^2=(a_1^2+〖2a_1 a〗_1+a_2^2 )=(b_1^2+〖2b_1 b〗_2+b_2^2 )
c=a_1+a_2,d= b_1+b_2,for c,d ∈R.So c^2=d^2.This is not the same as a^2 =b^2, so S isn't a subspace of R^2 ...?
(B)
(c) not a subspace..
In conclusion, I have only one subspace, which is part (a)...? Also, in (a)confusion: a+b =0 is R^2, not R^3...
I'd appreciate your great help as always. I'm studying this by myself and so no teacher I can ask for help.
Cheers.

a) I agree, it looks like a subspace, it represents the intersection of 2 non-parallel planes thourgh the origin in R^3
b) take (1,-1) clearly a^2=b^2 but a does not equal b
so take 2 elements in the set
(1,-1) and (1,1)
what is their addition?
c) well the zero function clearly isn't in the space...

(a)confusion: a+b =0 is R^2, not R^3...

(a,b,c) is an element of R^3
a+b=0 is a constraint only based on the first 2 elements

 

[mccoy1]

a) I agree, it looks like a subspace, it represents the intersection of 2 non-parallel planes thourgh the origin in R^3
b) take (1,-1) clearly a^2=b^2 but a does not equal b
so take 2 elements in the set
(1,-1) and (1,1)
what is their addition?
c) well the zero function clearly isn't in the space...


(a,b,c) is an element of R^3
a+b=0 is a constraint only based on the first 2 elements

Thanks for the help Lanedance.
Great, we both agree on part (a)-subspace of R^3, and (c)-not a subspace of P2.
Now back to (b)

b) take (1,-1) clearly a^2=b^2 but a does not equal b
so take 2 elements in the set
(1,-1) and (1,1)
what is their addition?

for point (1,-1), a^2 = b^2 but a isn't equal b.I agree.
Now let v1=(a1, b2) and v2=(a2, b2) be two vectors in S. v1+v2=(a1+a2, b1+b2) is also in S..but (a1+a2)^2=(b1+b2)^2 looks good when not expanded, but when you expand it: [a1^2+2(a1a2)+a2^2] =[b1^2+2(b1b2)+b2^2] then, it doesn't look good, I guess? What do you think?
Cheers.

 

[HallsofIvy]

Let u= <1, -1> and v= <1, 1>. Then u+ v= <2, 0> which is NOT in the set.

 

[mccoy1]

Let u= <1, -1> and v= <1, 1>. Then u+ v= <2, 0> which is NOT in the set.

I think the point you are putting across is that (a,b) isn't a subspace for R^2, given the condition that a^2 = b^2?
Cheers.