Subspaces in R4: Get Started & Understand Now

[Pengwuino]

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]

What are the axioms you need to check, and what can you verify and how?

 

[HallsofIvy]

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

 

[Pengwuino]

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]

What you need to do is to treat

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

as being synonymous with

"a + c = b + d"

 

[Pengwuino]

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

 

[Hurkyl]

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]

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?

 

[Pengwuino]

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?

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?

 

[Pengwuino]

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]

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.

 

[neurocomp2003]

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]

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.

let's do my example, since I refuse to answer your homework,

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.

so how about it, now?

You see how to do it?

 

[benorin]

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]

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]

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.