Is W a Subspace of All Ordered Pairs?

[winter_ken]

Homework Statement

Let W be the set of all ordereed pairs of real numbers, and consider the following addition and scalar multiplication operations on U=(u1,u2) and V=(v1,v2)

U+V is standard addition but kU=(0, ku2)

Homework Equations

Is W closed under scalar multiplication?

The Attempt at a Solution

I understand that W is not a vector space but my book suggests that it is a subspace closed by scalar multiplication.

Is it because kU=(k0, ku2) where k multiplies both terms?

 

[CompuChip]

Being closed under scalar multiplication just means that if $w \in W$ and $k \in \mathbb{R}$ then $k w \in W$.
Since W doesn't really have any restrictions, you only need to check that (0, ku₂) is an ordered pair of real numbers.

 

[Skins]

Is $$kU \, = \, (0, ku_{2})$$ an ordered pair of reals ? Also, does W contain the zero vector ?

 

[Fredrik]

It's pretty strange to talk about subspaces here, since W isn't defined as a subset of a vector space. It's just defined as a set with two operations which may or may not turn it into a vector space.

If it turns out to be closed under both addition and (this non-standard) scalar multiplication, then you can check if it satisfies the eight vector space axioms, to see if it's a vector space.

 

[winter_ken]

Thanks for the help! That solves it for me completely! I thoroughly didn't understand the definitions of subspace and what they mean and represent, so this simple explanation really clears up a big misconception that I had.