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U={(a,b,0) | a, b [tex]\in R[/tex]} and W={(a,b,c) | a+b+c=0, a,b,c [tex]\in R[/tex]}

U, W are vector subspaces from R[tex]\stackrel{3}{}[/tex].

P.S this is not homework question, just example, for my better understanding.

Thanks.

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- #1

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U={(a,b,0) | a, b [tex]\in R[/tex]} and W={(a,b,c) | a+b+c=0, a,b,c [tex]\in R[/tex]}

U, W are vector subspaces from R[tex]\stackrel{3}{}[/tex].

P.S this is not homework question, just example, for my better understanding.

Thanks.

- #2

JasonRox

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I don't understand what you're asking for.

You want to prove they're subspaces?

- #3

Tom Mattson

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Physicsissuef,

I'll just copy and paste my response to you from sciforums.

You've got to find the subspace that simultaneously satisfies the requirements for membership in

*U requires a zero 3rd coordinate.

*W requires the sum of the coordinates to be zero.

Taken together, we have:

[tex]U\cap W=\{(a,b,0)|a+b=0\}[/tex]

We can do even better. Since [itex]a+b=0[/itex] it follows that [itex]b=-a[/itex], and so:

[tex]U\cap W=span\{(1,-1,0)\}[/tex]

- #4

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And how will I know, which subspace requires, what?

Physicsissuef,

I'll just copy and paste my response to you from sciforums.

You've got to find the subspace that simultaneously satisfies the requirements for membership inbothof the subspaces.

*U requires a zero 3rd coordinate.

*W requires the sum of the coordinates to be zero.

Taken together, we have:

[tex]U\cap W=\{(a,b,0)|a+b=0\}[/tex]

We can do even better. Since [itex]a+b=0[/itex] it follows that [itex]b=-a[/itex], and so:

[tex]U\cap W=span\{(1,-1,0)\}[/tex]

- #5

Tom Mattson

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You are explicitly told which subspace requires what. Did you not read the problem you typed out?

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And in another problem, how will I recognize ?You are explicitly told which subspace requires what. Did you not read the problem you typed out?

- #7

Tom Mattson

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They will explicitly tell you in another problem, too. You just have to read it.

- #8

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*U requires a zero 3rd coordinate

*W requires a zero 1st coordinate

Taken together

Is it like this:

[tex]U\cap W=\{(0,z,0)|z [/tex] [tex]\in R[/tex]}

or

[tex]U\cap W=\{(0,y,0)|y [/tex] [tex]\in R[/tex]}

?

- #9

Tom Mattson

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"Yes" to both. Those are exactly the same.Is it like this:

[tex]U\cap W=\{(0,z,0)|z [/tex] [tex]\in R[/tex]}

or

[tex]U\cap W=\{(0,y,0)|y [/tex] [tex]\in R[/tex]}

?

- #10

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Ok, thanks.

- #11

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[itex]U=\{(a,b,0)|a=b, a,b [/itex] [itex]\in[/itex] R}

[itex]W=\{(0,b,c)|b=c, b,c [/itex] [itex]\in[/itex] R}

Is the result:

[tex]U\cap W= (0,b,0) | b[/tex] [tex]\in R}[/tex]

Will it be same for:

[itex]U=\{(a,b,0)| a,b [/itex] [itex]\in[/itex] R}

[itex]W=\{(0,b,c)|b,c [/itex] [itex]\in[/itex] R}

?

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- #13

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But is the mine correct? (The 1-st one)

- #14

HallsofIvy

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You have U= {(a, b, 0)|a= b} which I think would be better written {(a, a, 0)} and V= {(0, b, c)|b= c} which I think would be better written {(0, b, b)}. In order to be in the intersection, the vector must be in U which means the third component must be 0. In order to be in the V, the first component must be 0 and the second component must be the same as the third- which must be 0. The only vector in the intersection of U and V is the 0 vector: (0, 0, 0).

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[itex]U=\{(a,b,0)| a,b [/itex] [itex]\in[/itex] R}

[itex]W=\{(0,b,c)|b,c [/itex] [itex]\in[/itex] R}

Will [tex]U \cap W=\{0,b,0\}[/tex] ?

- #16

HallsofIvy

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You understand, don't you, that what they call the components doesn't matter. If the problem had said [itex]U=\{(a,b,0)| a,b\in R[/itex]} and [itex]W= \{(0, y, z)|y, z\in R[/itex]} Your answer could have been [itex]U \cap W=\{(a,b,0)\}[/itex] or [itex]U \cap W=\{(0,y,0)\}[/itex] or even [itex]U \cap W=\{(0,p,0)|p\in R\}[/itex]

- #17

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[itex]U=\{(0,y,z)|2y+3z=0\}[/itex] of [itex]\mathbb{R}^3[/itex]

[itex]W=\{(x,y,z)|x+y+z=0\}[/itex] of [itex]\mathbb{R}^3[/itex]

How we will find [tex]U\cap W[/tex]?

I mean, is it possible?

- #18

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yes it is possible, you have to solve the equations

x=0

2y+3z=0

z+y+x=0

x=0

2y+3z=0

z+y+x=0

- #19

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x=0yes it is possible, you have to solve the equations

x=0

2y+3z=0

z+y+x=0

y=0

z=0

Can somebody explain?

- #20

HallsofIvy

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[itex]U=\{(0,y,z)|2y+3z=0\}[/itex] of [itex]\mathbb{R}^3[/itex]

[itex]W=\{(x,y,z)|x+y+z=0\}[/itex] of [itex]\mathbb{R}^3[/itex]

How we will find [tex]U\cap W[/tex]?

I mean, is it possible?

yes it is possible, you have to solve the equations

x=0

2y+3z=0

z+y+x=0

In order to be in U, you must have x= 0 and 2y+ 3z= 0. In order to be in V, you must have z+ y+ x= 0. In order to be in their intersection, all three must be true- that's why mrandersdk to solve the three equations simultaneously. What you found is that the only vector that satisfies all three conditions and so the only vector in both U and V is the zero vector: <0, 0, 0>. That's not at all odd. The zero vector is inx=0

y=0

z=0

Can somebody explain?

- #21

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But isn't <0,0,0> dimension 1?In order to be in U, you must have x= 0 and 2y+ 3z= 0. In order to be in V, you must have z+ y+ x= 0. In order to be in their intersection, all three must be true- that's why mrandersdk to solve the three equations simultaneously. What you found is that the only vector that satisfies all three conditions and so the only vector in both U and V is the zero vector: <0, 0, 0>. That's not at all odd. The zero vector is ineverysubspace and so is in the intersection of any set of subspaces. The subspace containingonlythe zero vector is, also, the only subspace containing only one vector (indeed, the only subspace containing only a finite number of vectors) and the only subspace of dimension 0: the "trivial" subspace.

- #22

morphism

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Are you asking if the vector space consisting of only the zero vector is 1 dimensional?But isn't <0,0,0> dimension 1?

The answer is no, it's not. By definition a 1 dimensional vector space must have a basis consisting of one element. However, no basis can contain the zero vector, because any set that contains the zero vector is linearly dependent. So the zero vector space, since it doesn't contain any nonzero vectors, cannot actually have a basis that contains any elements. Conventionally, one would say that the empty set is then a basis for the zero vector space, and thus the zero vector space is 0 dimensional.

- #23

HallsofIvy

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Don't confuse (as you seem to be doing) the

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