Hi everybody....I really need some help. I did this assignment, but it got sent bac

[TrystanandEmma1]

I will put what I did....and the comments of the evaluator....if someone could step by step tell me what I did wrong and how to fix it...I will be forever in your debt.

Assignment:
Vector spaces possess a collection of specific characteristics and properties.

The set of elements belonging to R² is usually denoted as {(a, b) | a, b ∈ R}. Combining elements within this set under the operations of addition and scalar multiplication should use the following notation:

Define a nontrivial subspace of R², showing all work.

Here is what I did: In order for me to define a nontrivial subspace of , I need to first understand
what nontrivial actually is. Nontrivial is a linear equation in which the value of at least
one variable of the equation is not equal to 0. So, to describe a nontrivial subspace, I would need a linear equation in which one of the variables is not equal to 0. Therefore, let v=(5, 4) and u=(0, 0) in R^2. . In order for me to show that this is a subspace of R^2, I need to show three things. First, I need to show that 0 is in R^2) . Secondly, I would need to show closure for addition. Lastly, I need to show closure for scalar multiplication. Now, to show that 0 is in R^2, I need to first come up with a linear equation that goes through the origin and the point (5, 4). Therefore, my linear equation is y=5/4x . This is the linear equation that satisfies both points and goes through origin because I will plug in the points and see if I get equality. Therefore, 5= 5/4(4)=5 and 0= 5/4(0)=0. Therefore, v and u are in R^2. This shows that 0 is in R^2 . Now, I need to show that( u+v) is in R^2 . Now I will show that it is closed under addition: u+v=(4, 5)+(0, 0)=(4+0, 5+0)=(4, 5) which is in R^2 as shown above. Therefore, it is closed under addition. Now, we need to show that the set is closed under scalar multiplication. I am going to let my scalar=5, then 5u=5(4, 5)=(20, 25) which is in R^2 because 25=5/4(20) =25. Since, from above, I showed that 0 is in R^2 , u+v is in R^2, and c(u) is in R^2 , this would be a nontrivial subspace of R^2.

The comments I got when it was sent back to me is: A lot of discussion was provided on this part of the task. However, it is not clear as to what the non-trivial subspace is since two vectors were provided. Two vectors do not constitute a subspace.

Can anyone tell me what the grader meant by this comment and can someone tell me how I can fix it....thank you....I will be forever in your debt....

 

[daon2]

A subspace of R^2 which contains a non-zero vector will have infinitely many vectors in it, having two vectors is.... not enough. For example, if (5,4) belongs to your subspace, so must (10,8), (-5,-4), etc. It is not clear what your definition of nontrivial is (you are talking about equations?), but it typically means it is not {(0,0)} not R^2.

R^2 is a two dimensional vector space. Any proper subspace will be 1-dimensional. You have a solution in the middle of your paragraph. All points (x,y) satisfying y=5/4x.

 

[HallsofIvy]

I will put what I did....and the comments of the evaluator....if someone could step by step tell me what I did wrong and how to fix it...I will be forever in your debt.

Assignment:
Vector spaces possess a collection of specific characteristics and properties.

The set of elements belonging to R² is usually denoted as {(a, b) | a, b ∈ R}. Combining elements within this set under the operations of addition and scalar multiplication should use the following notation:

Define a nontrivial subspace of R², showing all work.

Here is what I did: In order for me to define a nontrivial subspace of , I need to first understand
what nontrivial actually is. Nontrivial is a linear equation

No, "nontrivial" is an adjective while "equation" is a noun so they can't be the same. A "nontrivial subspace" is a subspace that has more than one vector in it. It is possible, sometimes, to define a subspace using an equation.

in which the value of at least
one variable of the equation is not equal to 0. So, to describe a nontrivial subspace, I would need a linear equation in which one of the variables is not equal to 0. Therefore, let v=(5, 4) and u=(0, 0) in R^2. . In order for me to show that this is a subspace of R^2, I need to show three things. First, I need to show that 0 is in R^2) . Secondly, I would need to show closure for addition. Lastly, I need to show closure for scalar multiplication. Now, to show that 0 is in R^2, I need to first come up with a linear equation that goes through the origin and the point (5, 4). Therefore, my linear equation is y=5/4x . This is the linear equation that satisfies both points and goes through origin because I will plug in the points and see if I get equality. Therefore, 5= 5/4(4)=5 and 0= 5/4(0)=0. Therefore, v and u are in R^2. This shows that 0 is in R^2 . Now, I need to show that( u+v) is in R^2 . Now I will show that it is closed under addition: u+v=(4, 5)+(0, 0)=(4+0, 5+0)=(4, 5) which is in R^2 as shown above. Therefore, it is closed under addition. Now, we need to show that the set is closed under scalar multiplication. I am going to let my scalar=5, then 5u=5(4, 5)=(20, 25) which is in R^2 because 25=5/4(20) =25. Since, from above, I showed that 0 is in R^2 , u+v is in R^2, and c(u) is in R^2 , this would be a nontrivial subspace of R^2.

The comments I got when it was sent back to me is: A lot of discussion was provided on this part of the task. However, it is not clear as to what the non-trivial subspace is since two vectors were provided. Two vectors do not constitute a subspace.

Can anyone tell me what the grader meant by this comment and can someone tell me how I can fix it....thank you....I will be forever in your debt....

The difficulty appears to be that you have seen examples of subspaces but do not know what the definition is. A "nontrivial subspace" of a vector space, V, is a subset of V, containing the 0 vector and at least one non-zero vector that is "closed under addition" (if u and v are in the subset so is u+v) and is "closed under scalar multiplication (if v is in the subset and a is any scalar, then av is also in the subset. Your main error is that you show that if u and v are in the subset, then u+ v is in the space, R^2, not in the subset, and the same for scalar multiplication. If u= (4, 5) then 5u= 5(4, 5)= (20, 25) which is NOT in the set {(4, 5), (0, 0)}. In fact, since, if \(\displaystyle a\ne 1\), \(\displaystyle au\ne u\), any nontrivial subspace MUST contain an infinite number of vectors. Perhaps you mean to say that the subspace is the set of vectors, {(4x, 5x)}, for x any real number, but you did not say that.