- #1

Pengwuino

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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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- Thread starter Pengwuino
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- #1

Pengwuino

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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.

- #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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- #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"

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

as being synonymous with

"a + c = b + d"

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

- #8

matt grime

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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:

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

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[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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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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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

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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

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