Sarahisme Posted July 31, 2005 Share Posted July 31, 2005 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 Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 ok i've got a proof.... what do you guys think? Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 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? Link to comment Share on other sites More sharing options...
MetaFrizzics Posted July 31, 2005 Share Posted July 31, 2005 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? Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 haha yes anyways... well what about for part a) how do you think my answer for that looks? Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 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. Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 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 ? Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 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) Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 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 ? Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 oh ok , well i'll try again later and show you, thanks though Link to comment Share on other sites More sharing options...
Dave Posted July 31, 2005 Share Posted July 31, 2005 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). Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 cap 'n' cup, eh ? Thanks ! Link to comment Share on other sites More sharing options...
Sarahisme Posted August 1, 2005 Author Share Posted August 1, 2005 ok i am really stuck now, can someone please give me a hint? :S Link to comment Share on other sites More sharing options...
Sarahisme Posted August 1, 2005 Author Share Posted August 1, 2005 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) ??? ??? Link to comment Share on other sites More sharing options...
Dave Posted August 1, 2005 Share Posted August 1, 2005 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. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 2, 2005 Author Share Posted August 2, 2005 hang on i think i've got it...tell me if this is right, give me a min ... Link to comment Share on other sites More sharing options...
Sarahisme Posted August 2, 2005 Author Share Posted August 2, 2005 phew, here it is, what do you guys think now? dave? Link to comment Share on other sites More sharing options...
Dave Posted August 2, 2005 Share Posted August 2, 2005 That's okay, but you don't need to keep saying "by defn of a subspace". Link to comment Share on other sites More sharing options...
Sarahisme Posted August 2, 2005 Author Share Posted August 2, 2005 lol phew! finally got a proof right! yay! thanks dave Link to comment Share on other sites More sharing options...
DQW Posted August 2, 2005 Share Posted August 2, 2005 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. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 3, 2005 Author Share Posted August 3, 2005 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 Link to comment Share on other sites More sharing options...
Sarahisme Posted August 6, 2005 Author Share Posted August 6, 2005 oh and thanks dave too! Link to comment Share on other sites More sharing options...
Dave Posted August 6, 2005 Share Posted August 6, 2005 Not a problem Link to comment Share on other sites More sharing options...
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