# Vector Spaces

I had this question in my book asking me to show these things in detail, but it seems easy yet i dont understand why teacher said it was a little difficult:

1) Prove that R^2(with the rules of addition and scalar multiplication) is a vector space and find (zero vector)?

2)Deduce from the rules of addition and scalar multiplication that
0v=0(vector) for all v in V and (-1)v=-v? (v is a vector)

3) Show that a vector space U has another just 1 element or infinitely many.

I just need to see what you guys comment on this, as to me it seems like i just simply have to rewrite the rules and thats pretty much it, but again i dont think thats what its asking for, i just dont understand i guess.

Galileo
Homework Helper
Start with the definition of a vector space. Show that R^2 with the given rules of addition and scalar multiplication satisfies all the requirements of a vector space.

So rewriting the rules, that's pretty much it. You should only be careful to prove the required rules only from properties of operations in R and not R^2.

For b) and c) you must prove it from the defining properties of a vector space.

matt grime
Homework Helper
3) is false, but not if we mean "over R".

HallsofIvy
Homework Helper
matt grime said:
3) is false, but not if we mean "over R".

I had to think about that for a moment- it's false if the vector space is over a finite field!

2) just seems a simple application of the scalar multiple property. Start by defining the vector in some arbitrary way (v_1, v_2, ... , v_n) and use the property of real numbers, etc. to deduce what you need.

yea for 1) is this enough to prove that R^2 is a vector space:
The vector space of R^2 consists of the set of ordered real numbers
x = (x1, x2). Addition of these real numbers is defined by adding the components:
x + y = (x1 + y1, x2 + y2) = y + x.
And multiplication by a scalar k in R^2 is defined as
kx = (kx1, kx2).
(now what does it mean to find a zero vector?)
and for 3) i came up with, the zero vector as having one element and everything else besides zero vector in U, will have infinetly many elements. right?

matrix_204 said:
(now what does it mean to find a zero vector?)
Find a vector in R2 that behaves like the zero vector in the vector space axioms. You may think the choice is obvious, but you must supply the rigor.
matrix_204 said:
...and everything else besides zero vector in U, will have infinetly many elements...
Have you rigorously shown why they will have infinitely many elements (over the field R) ?

hypermorphism said:
Have you rigorously shown why they will have infinitely many elements (over the field R) ?

The set {0} is a subspace with just one element.
Every other subspace of U is infinite, so it has infinite number of elements, since 0 =/= v in U, then for any real number k, kv in U, since U is closed under scalar multiplication and for k ≠ d, kv ≠ dv. Proof of this is as follows:
If kv=dv, then (k – d)v = 0, so k – d ≠ 0, then
v = 1/(k – d) x (k – d)v = 1/(k – d) x 0 = 0.

Does this show everything or not?

matt grime