Vector subspace section

  • #1
Hi! I just want to ask you, what is the principle of finding section between two vector subspaces. Let's say:

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.
 

Answers and Replies

  • #2
JasonRox
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You want to find a section between two vector subspaces?

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

You want to prove they're subspaces?
 
  • #3
Tom Mattson
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He meant "intersection" (he posted the same thing at sciforums).

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 both of 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]
 
  • #4
He meant "intersection" (he posted the same thing at sciforums).

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 both of 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]
And how will I know, which subspace requires, what?
 
  • #5
Tom Mattson
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You are explicitly told which subspace requires what. Did you not read the problem you typed out?
 
  • #6
You are explicitly told which subspace requires what. Did you not read the problem you typed out?
And in another problem, how will I recognize ?
 
  • #7
Tom Mattson
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They will explicitly tell you in another problem, too. You just have to read it.
 
  • #8
And if U={(x,y,0) | x,y [tex]\in R[/tex] } and W={(0,z,t) | z,t [tex]\in R[/tex]}

*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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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]}
?
"Yes" to both. Those are exactly the same.
 
  • #11
And what if I have:
[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}

?
 
  • #12
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OK, you need to develop a taste, take one from U and plug it into W, so you have something like [tex] (a,b,0) [/tex] and you want this to satisfy the form of W so that [tex](a,b,0) = (0,b,c) [/tex] for some a,b,c. What do you think?
 
  • #13
OK, you need to develop a taste, take one from U and plug it into W, so you have something like [tex] (a,b,0) [/tex] and you want this to satisfy the form of W so that [tex](a,b,0) = (0,b,c) [/tex] for some a,b,c. What do you think?
But is the mine correct? (The 1-st one)
 
  • #14
HallsofIvy
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No, you are not correct. You ignored the "a= b" and "b= c" parts.

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).
 
  • #15
Ok, I understand now. And what for the 2-nd example.
[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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Yes. To be in U, the third component must be 0. To be in W, the first component must be 0. To be in both, the first and third components must be 0.

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
Ok, I have one more question. What if:
[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
 
  • #19
yes it is possible, you have to solve the equations

x=0
2y+3z=0
z+y+x=0
x=0
y=0
z=0
Can somebody explain?
 
  • #20
HallsofIvy
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Ok, I have one more question. What if:
[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
x=0
y=0
z=0
Can somebody explain?
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 in every subspace and so is in the intersection of any set of subspaces. The subspace containing only the 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.
 
  • #21
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 in every subspace and so is in the intersection of any set of subspaces. The subspace containing only the 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.
But isn't <0,0,0> dimension 1?
 
  • #22
morphism
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But isn't <0,0,0> dimension 1?
Are you asking if the vector space consisting of only the zero vector is 1 dimensional?

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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a one dimensional vector space could be modeled by any straight line passing through the origin. The only possible 0 dimensional space, as I said above, is the single 0 vector above.

Don't confuse (as you seem to be doing) the vector space with a basis for that vector space! The vector space having {<1, 1, 2>} as basis, for example, contains an infinite number of vectors- in this example, any vector of the form <a, a, 2a> where a can be any real number.
 

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