# Subspaces in R4

Gold Member
I'm so lost!

1. W is the set of all vectors in R4 such that x1 + x3 = x2 + x4. Is W a subspace of R4 and Why?

How do i get started here? i'm thoroughly confused on this whole idea of vector spaces and such.

matt grime
Homework Helper
What are the axioms you need to check, and what can you verify and how?

HallsofIvy
Homework Helper
I find looking at the definitions helpful in such a case.

Gold Member
So far we're looking at vector addition, scalar multiplication and the zero vector. Plugging in 0,0,0,0 produces a vector that is in the set. I'm not sure how to show vector addition however. Do i need to turn that equation into a function x1= x2 + x4 - x3?

Hurkyl
Staff Emeritus
Gold Member
What you need to do is to treat

"(a, b, c, d) is in W"

as being synonymous with

"a + c = b + d"

Gold Member
Can i set a new vector equal to like (y1 + y3 = y2 + y4) and add the equations?

Hurkyl
Staff Emeritus
Gold Member
No. You need to treat

"(a, b, c, d) is in W"

as being synonymous with

"a + c = b + d"

So far we're looking at vector addition, scalar multiplication and the zero vector.
So what, abstractly speaking, are you trying to show?

matt grime
Homework Helper
Take two elements that satisfy the condition, add them, does the result satisfy the condition?

suppose i wanted to show that the subset V of R^2 defined by (x,y) in V if x=0, then let (x,y) and (s,t) be in V, the sum is (x+s,y+t), now is x+s zero when x and s are zero?

Gold Member
Hurkyl said:
No. You need to treat

"(a, b, c, d) is in W"

as being synonymous with

"a + c = b + d"

Ok so how can I show that they add up using two different vectors? I know for example, (1,3,4,2) are in W but how do i go about showing that all such vectors satisify vector addition and multiplication?

Hurkyl said:
So what, abstractly speaking, are you trying to show?

I guess i'm trying to show that every vector that follows a + c = b + d is a vector in W. Is that right?

Gold Member
matt grime said:
Take two elements that satisfy the condition, add them, does the result satisfy the condition?

suppose i wanted to show that the subset V of R^2 defined by (x,y) in V if x=0, then let (x,y) and (s,t) be in V, the sum is (x+s,y+t), now is x+s zero when x and s are zero?

If I pick v = (1, 3, 4, 2) and u = ( 3, 6, 5, 2), they add up to be (4, 9, 9, 4) which is also in the set. I can also do cv, c being say, -3, to get cv = ( -3, -9, -12, -6) which is also in the set. How can i show that it works for all numbers?

benorin
Homework Helper
Gold Member
Formally, we have W is defined by

$$W=\left\{ (x_1,x_2,x_3,x_4)\in\mathbb{R}^4 : x_1 + x_3 = x_2 + x_4 \right\}$$

The requirement $x_1 + x_3 = x_2 + x_4$ may be written as $x_4 = x_1 - x_2 + x_3$ and notice how once $x_1, x_2, x_3$ are chosen, $x_4$ is determined, and so it would seem we only need three parameters to express points in W...

If you need more, post on that wise.

don't use numbers...use variables

you are given a condition x1+x3=x2+x4, someone above
said look at (a,b,c,d) and you yourself posted can i do x1=...
NOW combine those 2 concepts to get a general vector satisfying your given condition. Note you should ask yourself how many unknowns should remain in your general vector

matt grime
Homework Helper
i am going to state categorically, and i'm sorry to do so, that you should ignore benorin.

it's very easy, so stop making it difficult.

Define W in R^2 by (x,y) is in W if and only if x=0.

let's show it is a subspace.

1. (0,0) is in W, clearly since the first coordinate is zero

2. if (x,y) and (u,v) are in W the (x+u,y+v) is in W since x+u=0 because x=0 and u=0

3. if (x,y) is in W then so is t(x,y)=(tx,ty) since x=0 implies tx=0

and we are done.

You see how to do it?

benorin
Homework Helper
Gold Member
If $$\forall\alpha , \beta\in\mathbb{R},x,y\in W\Rightarrow (\alpha x+\beta y) \in W ,$$ then W is a subspace. That's it. Simple.

Parts (1), (2), and (3) of your test correspond to the cases $$\alpha=\beta=0, \alpha=\beta=1, \mbox{ and }\alpha=t,\beta=0,$$ respectfully.

matt grime
Homework Helper
benorin said:
The requirement $x_1 + x_3 = x_2 + x_4$ may be written as $x_4 = x_1 - x_2 + x_3$ and notice how once $x_1, x_2, x_3$ are chosen, $x_4$ is determined, and so it would seem we only need three parameters to express points in W...

i meant it was this comment that should be ignored. i don't see what choosing and determing things has to do with proving its a subspace

benorin
Homework Helper
Gold Member
I was attempting to lead to the representation of vectors in W as vectors of the form (x1,x2,x3,x1-x2+x3) is all.