Linear Algebra help please

[affini]

I have the following question to answer

Give an example of a subset of R² that is a nontrivial subspace of R², showing all work.

My definition of a nontrivial subspace: A non-trivial subspace of vector space, V, contains the zero vector and at least one non-zero vector, and is closed under addition and scalar multiplication. I used the equation y = (4/3)x

1. 0 is in R2

The equation goes through the origin and the point (3, 4)
so 4 = (4/3)*3 = 4
and 0 = (4/3)*0 = 0
2. closed under addition: <3, 4> + <0, 0> = <3, 4>
3. closed under scalar multiplication
If I make my scalar multiple = 2, then the result is:
2<3, 4> = <6, 8>
8 = (4/3)*6 = 8

I got this feedback: The set of vectors of the form (x,y) with y = (4/3)x is suggested as a subspace of R^2. The definition of a nontrivial subspace is almost correct. There is an additional condition that should be stated. The zero vector is shown to be in the set of vectors. The use of two specific vectors to establish that the suggested set of vectors is closed under addition and scalar multiplication is not adequate.

 

[stapel]

I have the following question to answer

Give an example of a subset of R² that is a nontrivial subspace of R², showing all work.

My definition of a nontrivial subspace: A non-trivial subspace of vector space, V, contains the zero vector and at least one non-zero vector, and is closed under addition and scalar multiplication. I used the equation y = (4/3)x

1. 0 is in R2

The equation goes through the origin and the point (3, 4)
so 4 = (4/3)*3 = 4
and 0 = (4/3)*0 = 0
2. closed under addition: <3, 4> + <0, 0> = <3, 4>
3. closed under scalar multiplication
If I make my scalar multiple = 2, then the result is:
2<3, 4> = <6, 8>
8 = (4/3)*6 = 8

I got this feedback: The set of vectors of the form (x,y) with y = (4/3)x is suggested as a subspace of R^2. The definition of a nontrivial subspace is almost correct. There is an additional condition that should be stated. The zero vector is shown to be in the set of vectors. The use of two specific vectors to establish that the suggested set of vectors is closed under addition and scalar multiplication is not adequate.

Check the definition of a subspace. You are missing one very important part of what it is to be a subspace, rather than merely a subset.

To prove closure, you need to show that any pair of vectors will work. You've only shown that those two particular vectors will work, and that's not enough. So start by defining what would be the form of generic vectors in this space. If y = (4/3)x, then wouldn't vectors be of the form <x, (4/3)x>? And, if so, where does this lead under the two operations?

 

[affini]

Check the definition of a subspace. You are missing one very important part of what it is to be a subspace, rather than merely a subset.

To prove closure, you need to show that any pair of vectors will work. You've only shown that those two particular vectors will work, and that's not enough. So start by defining what would be the form of generic vectors in this space. If y = (4/3)x, then wouldn't vectors be of the form <x, (4/3)x>? And, if so, where does this lead under the two operations?

I think I have it. I think I need to add that a subspace cannot be the original space.

 

[HallsofIvy]

I think I have it. I think I need to add that a subspace cannot be the original space.

What about the more important part of the problem? That you need to show that the sum of any two vectors in the subspace is in the subpace and you need to show that the product of any number with any vector in the subspace is in the subspace.

 

[affini]

What about the more important part of the problem? That you need to show that the sum of any two vectors in the subspace is in the subpace and you need to show that the product of any number with any vector in the subspace is in the subspace.

I got that too. I was just having issues with why the definition was wrong and then I realized why and was like, oh duh. lol