hi, yet another one of my questions

 

i can't quite figure out how to prove this....i can see its true, but i just don't know how to write it out formally :S

 

-Sez

ok i've got a proof.... what do you guys think?

oh and for part b) i would say something like

 

A =

[a]

[0]

 

B=

[0]

[c]

[d]

 

and the interesection of these two planes is a line along the y-axis

 

what do people reckon?

Two intersecting planes = a line. So that's good.

 

But does a circle count as a line?

Why not two touching circles in the same plane but different sizes?

haha yes anyways...

 

well what about for part a) how do you think my answer for that looks?

Two intersecting planes = a line. So that's good.

Not really. Any general plane does not constitute a vector space. But the specific planes chosen by Sarah will work.

 

But does a circle count as a line?

Why not two touching circles in the same plane but different sizes?

Please do not mislead students Meta.

1. Circles do not intersect at lines,

2. A circle can not make a vector space - it is closed under neither vector addition not scalar multiplication.

Sarah, I do not follow what you've done in part (a). What are v1, v2 ? Are they basis vectors of V ? And what are u and w, and why have you defined them to be null-vectors ?

v1 and v2 are vectors from V and u and w are vectors in the subspaces U and W respectively

 

or i suppose you could just call u and w the zero vector for each subspace (but since they are subspaces it is the same zero vector in both of the subspaces and the vector space V)

No Sarah, that proof is incorrect.

 

1. You are specifically trying to prove a result for a 2-dimensional vector space, instead of for any general vector space

 

2. You have not used anywhere that a and b actually belong in U/\W.

 

As always, start from the definitions :

[math]x~\epsilon~U~ int~ W \implies x ~ \epsilon~ U~and ~x~\epsilon~ W [/math] and conversely.

 

PS : What are the [imath]\LaTeX[/imath] codes for union and intersection, anyone ?

oh ok , well i'll try again later and show you, thanks though

In regards to the LaTeX commands, you can use [math]U \cap V[/math] and [math]U \cup V[/math]. (Click on the images for the LaTeX).

cap 'n' cup, eh ? Thanks !

ok i am really stuck now, can someone please give me a hint? :S

i can see that UnW must contain the zero vector that is in the vector space V, but this is only one of the three properties that need to be satified, and i am not sure how to show the other two properties (closed under addition and closed under multiplication)

???

???

Think about it. Take an element v in U int W and a constant k in the associated field. What do you know about that element? Well, for sure, v is in U and also, v is in W. Now, since U is a vector field, surely kv is in U as well?

 

This is such a big hint that you should be able to get it from here. The proof is literally 3 lines long - one for each of the cases.

hang on i think i've got it...tell me if this is right, give me a min ...

phew, here it is, what do you guys think now? dave?

That's okay, but you don't need to keep saying "by defn of a subspace".

lol

 

phew! finally got a proof right! yay!

 

thanks dave

The last 2 places where you write "by definition of a subspace", you should write (if you want to write anything at all) "by definition of the intersection", because that's what you're using.

The last 2 places where you write "by definition of a subspace", you should write (if you want to write anything at all) "by definition of the intersection", because that's what you're using.

 

yep i see that, i agree , thanks DQW

oh and thanks dave too!

Not a problem