Linear Algebra - Vector Subspaces

[steelphantom]

Homework Statement

Give an example of a nonempty subset U of R² such that U is closed under scalar multiplication, but U is not a subspace of R².

Homework Equations

The Attempt at a Solution

I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R² because 0 is not in this subset.

Thanks!

 

[NateTG]

The Attempt at a Solution

I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R² because 0 is not in this subset.

You get $<0,0>$ if you multiply by $a=0$.
(Hint: What are the properties of subspaces?)

 

[steelphantom]

You get $<0,0>$ if you multiply by $a=0$.
(Hint: What are the properties of subspaces?)

1) Additive Identity: 0 is in U.

2) Closed Under Addition: u, v in U implies u + v is in U.

3) Closed Under Scalar Multiplication: a is in R (or C) and u is in U implies au is in U.

So I need to come up with a subset that fails for one (or both) of the first two properties, but not for property 3.

How about checking for property 2? If u = (x, x + 2) and v = (y, y + 2), where u, v are in U, then u + v = (x + y, (x + y) + 4), which is not in U, so U is not a subspace of R².

 

[NateTG]

How about checking for property 2? If u = (x, x + 2) and v = (y, y + 2), where u, v are in U, then u + v = (x + y, (x + y) + 4), which is not in U, so U is not a subspace of R².

...better, but:
$$2<3,(3+2)>=<6,10> \notin U$$
so you don't have closure under scalar multiplication.

 

[steelphantom]

...better, but:
$$2<3,(3+2)>=<6,10> \notin U$$
so you don't have closure under scalar multiplication.

Wow, I guess I don't. I really don't know where to go from here. Any hints on a a type of subset that would work in this situation?

 

[steelphantom]

I was searching the forums and found pretty much the exact same question. Here's the thread if anyone is interested: https://www.physicsforums.com/showthread.php?t=41884"

A set that works is the following: U = {(x, y) | xy = 0}. This set is closed under scalar multiplication, but not vector addition.

 

[studguy]

Homework Statement

Give an example of a nonempty subset U of R² such that U is closed under scalar multiplication, but U is not a subspace of R².

Homework Equations

The Attempt at a Solution

I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R² because 0 is not in this subset.

Thanks!

You are absolutely wrong.U={(x,x+2)| x is in R} is not closure under multiplication.
Ex:put x=0,a=2 (0,2) is in U but (0,4) is not in U.

 

[HallsofIvy]

You are absolutely wrong.U={(x,x+2)| x is in R} is not closure under multiplication.
Ex:put x=0,a=2 (0,2) is in U but (0,4) is not in U.

Yes. he was told that almost two years ago!