Subspaces in R4

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  • #1
Pengwuino
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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.
 

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  • #2
matt grime
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What are the axioms you need to check, and what can you verify and how?
 
  • #3
HallsofIvy
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I find looking at the definitions helpful in such a case.
 
  • #4
Pengwuino
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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?
 
  • #5
Hurkyl
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What you need to do is to treat

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

as being synonymous with

"a + c = b + d"
 
  • #6
Pengwuino
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Can i set a new vector equal to like (y1 + y3 = y2 + y4) and add the equations?
 
  • #7
Hurkyl
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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?
 
  • #8
matt grime
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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?
 
  • #9
Pengwuino
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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?
 
  • #10
Pengwuino
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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?
 
  • #11
benorin
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Formally, we have W is defined by

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

The requirement [itex]x_1 + x_3 = x_2 + x_4 [/itex] may be written as [itex]x_4 = x_1 - x_2 + x_3 [/itex] and notice how once [itex]x_1, x_2, x_3 [/itex] are chosen, [itex]x_4[/itex] 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.
 
  • #12
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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
 
  • #13
matt grime
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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?
 
  • #14
benorin
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If [tex]\forall\alpha , \beta\in\mathbb{R},x,y\in W\Rightarrow (\alpha x+\beta y) \in W ,[/tex] then W is a subspace. That's it. Simple.

Parts (1), (2), and (3) of your test correspond to the cases [tex]\alpha=\beta=0, \alpha=\beta=1, \mbox{ and }\alpha=t,\beta=0,[/tex] respectfully.
 
  • #15
matt grime
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benorin said:
The requirement [itex]x_1 + x_3 = x_2 + x_4 [/itex] may be written as [itex]x_4 = x_1 - x_2 + x_3 [/itex] and notice how once [itex]x_1, x_2, x_3 [/itex] are chosen, [itex]x_4[/itex] 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
 
  • #16
benorin
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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.
 

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