Vector subspace section

1. Feb 18, 2008

Physicsissuef

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 $$\in R$$} and W={(a,b,c) | a+b+c=0, a,b,c $$\in R$$}

U, W are vector subspaces from R$$\stackrel{3}{}$$.

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

2. Feb 18, 2008

JasonRox

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. Feb 18, 2008

Tom Mattson

Staff Emeritus
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:

$$U\cap W=\{(a,b,0)|a+b=0\}$$

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

$$U\cap W=span\{(1,-1,0)\}$$

4. Feb 18, 2008

Physicsissuef

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

5. Feb 18, 2008

Tom Mattson

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

6. Feb 18, 2008

Physicsissuef

And in another problem, how will I recognize ?

7. Feb 18, 2008

Tom Mattson

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

8. Feb 18, 2008

Physicsissuef

And if U={(x,y,0) | x,y $$\in R$$ } and W={(0,z,t) | z,t $$\in R$$}

*U requires a zero 3rd coordinate

*W requires a zero 1st coordinate

Taken together

Is it like this:
$$U\cap W=\{(0,z,0)|z$$ $$\in R$$}
or
$$U\cap W=\{(0,y,0)|y$$ $$\in R$$}
?

9. Feb 18, 2008

Tom Mattson

Staff Emeritus
"Yes" to both. Those are exactly the same.

10. Feb 18, 2008

Physicsissuef

Ok, thanks.

11. Feb 19, 2008

Physicsissuef

And what if I have:
$U=\{(a,b,0)|a=b, a,b$ $\in$ R}
$W=\{(0,b,c)|b=c, b,c$ $\in$ R}
Is the result:
$$U\cap W= (0,b,0) | b$$ $$\in R}$$

Will it be same for:

$U=\{(a,b,0)| a,b$ $\in$ R}
$W=\{(0,b,c)|b,c$ $\in$ R}

?

12. Feb 19, 2008

trambolin

OK, you need to develop a taste, take one from U and plug it into W, so you have something like $$(a,b,0)$$ and you want this to satisfy the form of W so that $$(a,b,0) = (0,b,c)$$ for some a,b,c. What do you think?

13. Feb 19, 2008

Physicsissuef

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

14. Feb 20, 2008

HallsofIvy

Staff Emeritus
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. Feb 20, 2008

Physicsissuef

Ok, I understand now. And what for the 2-nd example.
$U=\{(a,b,0)| a,b$ $\in$ R}
$W=\{(0,b,c)|b,c$ $\in$ R}
Will $$U \cap W=\{0,b,0\}$$ ?

16. Feb 20, 2008

HallsofIvy

Staff Emeritus
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 $U=\{(a,b,0)| a,b\in R$} and $W= \{(0, y, z)|y, z\in R$} Your answer could have been $U \cap W=\{(a,b,0)\}$ or $U \cap W=\{(0,y,0)\}$ or even $U \cap W=\{(0,p,0)|p\in R\}$

17. Feb 23, 2008

Physicsissuef

Ok, I have one more question. What if:
$U=\{(0,y,z)|2y+3z=0\}$ of $\mathbb{R}^3$
$W=\{(x,y,z)|x+y+z=0\}$ of $\mathbb{R}^3$
How we will find $$U\cap W$$?
I mean, is it possible?

18. Feb 23, 2008

mrandersdk

yes it is possible, you have to solve the equations

x=0
2y+3z=0
z+y+x=0

19. Feb 23, 2008

Physicsissuef

x=0
y=0
z=0
Can somebody explain?

20. Feb 23, 2008

HallsofIvy

Staff Emeritus
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.