Linear subspace constraints and notation.

chinye11
Messages
22
Reaction score
0

Homework Statement


The original printed problems can be found as attachments. The questions ask if a set S is a subset Rn. Give Reasons

Question 1.) S is the set of all vectors [x1,x2] such that x12 + x22 < 36Question 2.) S is the set of all vectors [x1,x2,x3] such that:
x2= 2x1
x3 = 3x1

Homework Equations



Definition of subspace: A subspace S of a vector space T is a subset which contains the zero vector and is closed under the vector addition and scalar multiplication which define T.
i.e. S is a subset of T which is also a vector space under the same operations.

The Attempt at a Solution


Ok, I thought that neither of these were subspaces of R3 so my reasoning is as follows:

Question 1) S is not a subspace since it is not a subset and is also not closed under vector addition or scalar multiplication.

Not closed under vector addition because:
x1 = 5 , x2 =1, would be an element of S but
[x1,x2]+[x1,x2] = [10,2] =2[x1,x2] is not an element of S.

I thought S was not a subset, since I can choose any complex numbers x1 =(a+bi) x2 = (c+di) such that the following values are true and they will be contained within the set S:
a2 -b2 +4 c2 - 4d2 < 36
2ab +8cd = 0

So as a simple example, x1 = 5i x2 =0.

Question 2.) S is not a subspace. This one is closed under vector addition and scalar multiplication and contains the zero vector. Again the constraint allows for the use of any complex numbers as far as I can tell and hence is not a subspace since it is not a subset of r3

I was in contact with my lecturer and he said that it is implied in the question {x1,x2} are elements of R2. I e-mailed him back asking how this implication came about only to find an out of office for the next month. So I was wondering if there was something in the notation used for these questions which confines it to the Real numbers as written above. This would change the answer to question 2 and give one less reason in question 1.

Just to clarify the information in the thumbnails is the full amount of information provided, there is no limiting to Real vector spaces given with instructions to the questions or any such limiting and we considered complex vector spaces in our course.
 

Attachments

  • Linear Question 1.jpg
    Linear Question 1.jpg
    13.1 KB · Views: 578
  • Linear Question 2.jpg
    Linear Question 2.jpg
    13 KB · Views: 820
Physics news on Phys.org
chinye11 said:

Homework Statement


The original printed problems can be found as attachments. The questions ask if a set S is a subset Rn. Give Reasons

Question 1.) S is the set of all vectors [x1,x2] such that x12 + x22 < 36Question 2.) S is the set of all vectors [x1,x2,x3] such that:
x2= 2x1
x3 = 3x1

Homework Equations



Definition of subspace: A subspace S of a vector space T is a subset which contains the zero vector and is closed under the vector addition and scalar multiplication which define T.
i.e. S is a subset of T which is also a vector space under the same operations.

The Attempt at a Solution


Ok, I thought that neither of these were subspaces of R3 so my reasoning is as follows:

Question 1) S is not a subspace since it is not a subset and is also not closed under vector addition or scalar multiplication.

Not closed under vector addition because:
x1 = 5 , x2 =1, would be an element of S but
[x1,x2]+[x1,x2] = [10,2] =2[x1,x2] is not an element of S.

I thought S was not a subset, since I can choose any complex numbers x1 =(a+bi) x2 = (c+di) such that the following values are true and they will be contained within the set S:
a2 -b2 +4 c2 - 4d2 < 36
2ab +8cd = 0

So as a simple example, x1 = 5i x2 =0.

Question 2.) S is not a subspace. This one is closed under vector addition and scalar multiplication and contains the zero vector. Again the constraint allows for the use of any complex numbers as far as I can tell and hence is not a subspace since it is not a subset of r3

I was in contact with my lecturer and he said that it is implied in the question {x1,x2} are elements of R2. I e-mailed him back asking how this implication came about only to find an out of office for the next month. So I was wondering if there was something in the notation used for these questions which confines it to the Real numbers as written above. This would change the answer to question 2 and give one less reason in question 1.

Just to clarify the information in the thumbnails is the full amount of information provided, there is no limiting to Real vector spaces given with instructions to the questions or any such limiting and we considered complex vector spaces in our course.

No. Both of your sets S are subsets of ##R^n##. What makes you think otherwise? (I suspect that there is a typo in the first question, and that it should say ##R^2## rather than ##R^3##. To be in a subset of ##R^3## the vector would need a third component.)

The issue is not whether they are subsets; it is whether they are subspaces, and that is a different matter.

So, is your first set S closed under addition? Under scalar multiplication? Ditto for your second S.

I see that you have more-or-less dealt with all these issues, but you come across as being a little bit confused.
 
Last edited:
Ray Vickson said:
No. Both of your sets S are subsets of ##R^n##. What makes you think otherwise? (I suspect that there is a typo in the first question, and that it should say ##R^2## rather than ##R^3##. To be in a subset of ##R^3## the vector would need a third component.)

The issue is not whether they are subsets; it is whether they are subspaces, and that is a different matter.

So, is your first set S closed under addition? Under scalar multiplication? Ditto for your second S.

I see that you have more-or-less dealt with all these issues, but you come across as being a little bit confused.

Thank you, Regarding the subset not subspace point. I am not sure about this since all the definition I can find for subspace say
"given a subset V of a vector space W", albeit maybe with different letters,
" V is a subspace of W if: 0 ε V

if vectors {x,y} are in V then x + y is also in V

if x is a vector in V and a is a scalar then ax is in V for all x ε V and scalars defined in the field."

Does the "given a subset" point mean I should presume my set is restricted to those values which will for a subset of desired vector space in this case R2.

So my issue arises with the following. If i say S is the set of all vectors of them form [x1,x2] such that x12 + x22 < 36. I can see instantly that there are vectors in S e.g. (6i,1) which satisfy the constraint but which are not elements of R2. To me this would suggest that S is not a subspace of R2.

I understand that this particular subset is not closed under either vector addition or scalar multiplication but to demonstrate my problem say I change the constraint to x12 + x22 ε R.

Now this would mean S would be closed under vector addition and scalar multiplication and contain the zero vector for the given operations. I understand S is a subset or equal to R if {x1,x2} ε R and these are the values of S which the later tests for closure and the zero value will test. However as far as I can tell (5i,0) ε S but (5i,0) is not an element of R2. Hence I thought S was not an subspace of R2.

Just to add an example to help me locate my problem since I find it difficult to word.

Are the set of complex numbers in one dimension under vector addition and real scalar multiplication a subspace of the real numbers in one dimension. I would argue no since the complex numbers are not a subset of the real numbers.

I am unsure of the effect of the "given a subset" terminology as regards what is allowed.

However as far as the other conditions go, the zero vector is in the complex numbers and a real number under vector addition with another real number or multiplication by a real scalar remains in the real numbers and therefore these conditions are satisfied.
 
Last edited:
chinye11 said:
Thank you, Regarding the subset not subspace point. I am not sure about this since all the definition I can find for subspace say
"given a subset V of a vector space W", albeit maybe with different letters,
" V is a subspace of W if: 0 ε V

if vectors {x,y} are in V then x + y is also in V

if x is a vector in V and a is a scalar then ax is in V for all x ε V and scalars defined in the field."

Does the "given a subset" point mean I should presume my set is restricted to those values which will for a subset of desired vector space in this case R2.

So my issue arises with the following. If i say S is the set of all vectors of them form [x1,x2] such that x12 + x22 < 36. I can see instantly that there are vectors in S e.g. (6i,1) which satisfy the constraint but which are not elements of R2. To me this would suggest that S is not a subspace of R2.

I understand that this particular subset is not closed under either vector addition or scalar multiplication but to demonstrate my problem say I change the constraint to x12 + x22 ε R.

Now this would mean S would be closed under vector addition and scalar multiplication and contain the zero vector for the given operations. I understand S is a subset or equal to R if {x1,x2} ε R and these are the values of S which the later tests for closure and the zero value will test. However as far as I can tell (5i,0) ε S but (5i,0) is not an element of R2. Hence I thought S was not an subspace of R2.

Just to add an example to help me locate my problem since I find it difficult to word.

Are the set of complex numbers in one dimension under vector addition and real scalar multiplication a subspace of the real numbers in one dimension. I would argue no since the complex numbers are not a subset of the real numbers.

I am unsure of the effect of the "given a subset" terminology as regards what is allowed.

However as far as the other conditions go, the zero vector is in the complex numbers and a real number under vector addition with another real number or multiplication by a real scalar remains in the real numbers and therefore these conditions are satisfied.

Some subsets are subspaces, and some subsets are not subspaces.
 
I want to try a different approach here since I am having difficulty expressing my problem.

Take the set of all real numbers in 1 dimension, i.e. R. Now say I add the value i to the set and name it Ri. Is this new set Ri a subspace of R?

In other words is it a requirement must a subspace be fully contained within i.e. subset or equal to it's vector space.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top