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/\/\isanThr0p
Nov 16th, 2003, 09:24 AM
hey everyone well here's my prob:
K is a Body, V is a vector space over K
u,v,w are element of V
Prove that the following statements are equivalent:
(i) u,v,w are linearly dependent
(ii) One of the following statements is true
(1) v=0
(2) a element of K with w=a*v is existing
(3) b,c element of K with b*v+c*w=u are existing
alright, I get how any of the (ii) statements imply (i), but not the other way round.... I hope anyone can help
sorry for my bad math english... if anyone can paraphrase it in the correct terms, he is more than welcome.
thx in advance
TheManWhoCan
Nov 16th, 2003, 02:43 PM
OK I gotta assume you mean u, v and w are perpendicular (all at right angles to each other), otherwise you could express one in terms of the other two and so one of them would be obsolete.
Well, start with the statements in (ii):
1: V = 0, clearly rubbish because then you wouldn't have a vector space at all - u, v, and w can't be perpendicularif they don't exist.
2: an element of K with w = av exists, also rubbish because that's saying that w is a scalar multiple of v, and if w is in the same direction as v it isn't perpendicular.
3: hmm that's also crap because it says you can make u out of scalar multiples of v and w, so they can't all be perpendicular.
What kind of vector space is this, that has one of it's defined variables in terms of the others?
/\/\isanThr0p
Nov 16th, 2003, 03:38 PM
yeah but I was asking the other way round in the first place so knowing the vectors are linearly dependent.
Also you don't need perpendicular vectors to make a space...
this can be shown in a plane (it's simpler in a plane)
let's say you have the vector (1,0) and (0,1)... with those you can obviously reach any point on the plane and they are perpendicular but you could also archive this with let's say (2,1) and (0,1)
for example (33,13)=10*(2,1)+13(0,1)
you could also prove it, but I think you got me
TheManWhoCan
Nov 16th, 2003, 04:05 PM
Yeh I'm just confused as to what you mean by an element of K, and linearly dependant.
/\/\isanThr0p
Nov 17th, 2003, 09:00 AM
it doesn't matter anyways since I had to send the problem in 2 days ago, but well I am not sure if body is the correct term.
linearly dependent is though. Here's an explanation:
u,v,w are vectors
0 is a vector containing only zeros (in R³ that is, in other vektorspaces the zero element)
a,b,c element of R
0=a*u*+b*v+c*w
if you can solve this equation with any other solution than a=b=c=0, the vectors are linearly dependent.
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