Determining if sets are in the subspace

[Renzokuken]

Homework Statement

is B in subspace R^2
B=[x] :x^2+y^2<=1
[y]

Homework Equations

1.0∈B
2.if u,v∈B u+v∈B
3.if u∈A a∈B then au∈B

The Attempt at a Solution

The <= is confusing me. I am not sure if i am suppose to treat it like an equal sign, or is it automatically not in the subspace?

if i treat it like a equal sign
0^2+0^2=0
u^2+uv+v^2 + u^2+uv+v^2 = 0
0+0=0
(au)^2+(au)^2=0
0^2+0^2=0
so B is in R^2

 

[HallsofIvy]

Homework Statement

is B in subspace R^2
B=[x] :x^2+y^2<=1
[y]

No, that makes no sense. You are asked to determine whether or not B is a subspace of R².

Homework Equations

1.0∈B
2.if u,v∈B u+v∈B
3.if u∈A a∈B then au∈B

Please do not use "special symbols"; they don't show up on all net-readers. Either use LaTex like 0\in B or write
1. 0 is in B.
2. if u and v are in B, then u+ v is in V
3. if a is in R and u is in B then au is in B.
Notice (3). What you wrote doesn't make sense because you did not say what "A" is. Also, if you have already used "u" to mean a vector, don't use it again to mean a number.
Finally, you don't need to prove (1). If (3) is true for any number a, take a= 0.

The Attempt at a Solution

The <= is confusing me. I am not sure if i am suppose to treat it like an equal sign, or is it automatically not in the subspace?

Why would that confuse you? it means "all pairs (x,y) such that x²+ y² is less than or equal to 1. And why would you think it is either "an equal sign" or "automatically not in the subspace"?

if i treat it like a equal sign
0^2+0^2=0

Which is NOT equal to 1 is it? So (0,0) would not be in this set. If this were only "=" this would immediately NOT be a subspace of R². But because 0< 1 and the problem DOES say "<= 1", you don't know yet.

u^2+uv+v^2 + u^2+uv+v^2 = 0

What? Where did you get this? I thought for moment you were doing (u+ v)²+ (u+ v)² but that would not give you this. And why 0 on the right?

0+0=0
(au)^2+(au)^2=0
0^2+0^2=0
so B is in R^2

No! if u is a vector, then u= (x, y) and au= (ax, ay). You want to show that (ax)²+ (ay)²<= 1, not (au)^2+ (au)^2! In a vector space only addition of vectors and multiplication of vectors by numbers is defined, not multiplication of two vectors and so (au)² makes no sense.

 

[Mark44]

I don't think you have given us the problem's exact wording. I believe that the problem's wording is something like this:
Let B = {(x, y): x^2 + y^2 <= 1}
Is B a subspace of R^2?

Please take a closer look at the problem as defined in your book.

You are confused about what you are supposed to do. Apparently you are trying to determine whether vectors are "in the subspace." What you need to do is show that the set B, which is a subset of R^2, actually is or is not a subspace of R^2.

Clearly the vector (0, 0) is in B.
If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
If the answer to each of these two questions is yes, no matter which vectors a and b and scalar c are chosen, then set B is a subspace of R^2.

If the answer to at least one question is no, then B is not a subspace of R^2.

 

[Renzokuken]

If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
a=x1^2+y1^2<=1
0^2+0^2<=1
b=x2^2+y2^2=1
0^2+0^2<=1
a+b<=1

If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
a^2(x^2+y^2)<=1
if x and y are equal to 0 or a=0, then it would be true.

so B is in the subspace of R^2?

 

[Mark44]

If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
a=x1^2+y1^2<=1
0^2+0^2<=1
b=x2^2+y2^2=1
0^2+0^2<=1
a+b<=1

If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
a^2(x^2+y^2)<=1
if x and y are equal to 0 or a=0, then it would be true.

so B is in the subspace of R^2?

You're still not getting it. The question is not whether B is in the subspace of R^2, but whether B ***is*** a subspace of R^2.

Suppose a and b are vectors in set B, where a = (x1, y1) and b = (x2, y2).
a in B ==> x1^2 + y1^2 <= 1, and
b in B ==> x2^2 + y2^2 <= 1.

Now, is a + b in B?
Also, is ca in B, where c is a scalar?

If the answers to both questions are "yes", then B is a subspace of R^2.

Think about what B looks like, and what a vector that belongs to B looks like.

 

[HallsofIvy]

If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
a=x1^2+y1^2<=1
0^2+0^2<=1

Why are you writing this? You cannot assume that x1 and y1 are 0.

b=x2^2+y2^2=1
0^2+0^2<=1

Same comment

a+b<=1

This makes no sense. a and b are vectors. You can't compare them to a number.

If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
a^2(x^2+y^2)<=1
if x and y are equal to 0 or a=0, then it would be true.

But x, y, and a are NOT necessarily 0!

so B is in the subspace of R^2?

You seem to be completely misunderstanding this problem.

 

[Renzokuken]

You're still not getting it. The question is not whether B is in the subspace of R^2, but whether B ***is*** a subspace of R^2.

Suppose a and b are vectors in set B, where a = (x1, y1) and b = (x2, y2).
a in B ==> x1^2 + y1^2 <= 1, and
b in B ==> x2^2 + y2^2 <= 1.

Now, is a + b in B?
Also, is ca in B, where c is a scalar?

If the answers to both questions are "yes", then B is a subspace of R^2.

Think about what B looks like, and what a vector that belongs to B looks like.

I think i am starting to understand it more.
A(x1^2+x2^2)+B(y1^2+y2^2)<=1 where A and B are scalers
a+b is not in B

ca is not in B either since c can be a scaler that makes a not a subspace in B
so B is not a subspace of R^2

 

[HallsofIvy]

I think i am starting to understand it more.
A(x1^2+x2^2)+B(y1^2+y2^2)<=1 where A and B are scalers
a+b is not in B

Assuming a= (x1,y1) and b= (x2,y2), this is Aa+ Bb. Now how do you conclude it is NOT in B? Do you mean there exist some a, b, A, B such that it is not in B or it is not in B for ALL a, b, A, B?

(If you are going to use "a" and "b" for vectors it is generally not a good idea to use "A" and "B" for scalars.)

ca is not in B either since c can be a scaler that makes a not a subspace in B
so B is not a subspace of R^2

You can't just say "ca is not in B", you have to show that. Can you give an example of a vector a in B and a number c such that ac is not in B?

 

[Renzokuken]

There exist some a ,b, A, B such that it is not in B.
a+b is (x1^2+x2^2+y1^2+y2^2)<=1, which is not always true.

ca is not in B because
a=(.5,.5) and c=10
100(.5^2+.5^2) is greater then 1 so ca is not in B.

 

[Renzokuken]

Is this enough proof to show that B is not a subspace of R^2?

 

[Mark44]

There exist some a ,b, A, B such that it is not in B.
a+b is (x1^2+x2^2+y1^2+y2^2)<=1, which is not always true.

ca is not in B because
a=(.5,.5) and c=10
100(.5^2+.5^2) is greater then 1 so ca is not in B.

For the first part, you have to SHOW that if a is in B and b is in B, then a + b isn't in B. You're saying it is not always true. Show us a case where it isn't true.

Your argument in the second part is more convincing - I truly believe that a is in B, but ca is not in B.