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Vector space explaination

  1. Sep 2, 2010 #1
    Hi,

    could you please tell me the collection of 3- vectors [a,b,c] form a vector space such that c - a =2b . Could you please explain it?

    thank you so much.
     
  2. jcsd
  3. Sep 2, 2010 #2

    Office_Shredder

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    It's not clear to me what your post is saying.

    Are you asking for an explanation on why the set of vectors {(a,b,c) such that c-a=2b} is a vector space?
     
  4. Sep 2, 2010 #3
    Hi Office_shredder,

    exactly..could you please explain?
     
  5. Sep 2, 2010 #4
    To decide whether something is a vector space, you need to see whether it complies with the vector space axioms. Here is one statement of them. I assume your vectors of the form (a,b,c) are sequences of three real numbers, called the components of the vector, with addition defined componentwise, that is: (x1, x2, x3) + (y1, y2, y3) = (x1 + y1, x2 + y2, x3 + y3), and scalar multiplication defined as multiplication of each component by a scalar, s, in this case a real number: s(a,b,c) = (sa, sb, sc). In other words, you want to check that your set makes a subspace of R3, and thus a vector space in its own right. Most of the axioms are fulfilled by these definitions, but you still need to answer the following questions. Is the zero vector, (0,0,0) part of this set? For all scalars (real numbers), s, does s(a,b,c) belong to this set? Does the sum of every pair of vectors in this set also belong to the set?
     
  6. Sep 2, 2010 #5
    Hi rasal,

    thanks for the reply. But how do I make sure that c-a=2b?

    thats where I get confused.

    thanks a lot.
     
  7. Sep 2, 2010 #6
    Its an arbitrary equation. It could be c+5a=7b or -3c+7a+6b=0 as well. There is something missing..
     
  8. Sep 2, 2010 #7
    woodyallen,

    thats exactly my question...

    for this specific equation, how do I prove that a,b,c form a vector space in such a way that c -a =2b?

    thanks
     
  9. Sep 2, 2010 #8
    Write the vector in the form (a,(c-a)/2,c). In order to form a vector space, the sum two of these must give a third of the same space(form) and a linear combination in general gives a third which belong in the same space..
     
  10. Sep 2, 2010 #9
    If you prove the linear combination you are ok..
     
  11. Sep 2, 2010 #10
    woodyallen,

    How do i prove the linear combination? Could you please elaborate? you draw me towards mathematics :)
     
  12. Sep 2, 2010 #11
    Suppose u1=(a1,(a1-c1),c1) and u2=(a2,(a2-c2),c2) i.e. each one of these fulfils the given relation, that is, it belongs to the vector space. Now uou have to prove that λu1 + κu2 belongs to vector space too. Replace u1 and u2 in the latter and you will find out that you get another vector that belongs to the same vector space since its coordinates are in accordance with the given equation.
     
  13. Sep 2, 2010 #12
    1 and 2 are subscripts and λ and κ scalars (numbers).
     
  14. Sep 2, 2010 #13
    I forgot a /2 at the second coordinate..
     
  15. Sep 2, 2010 #14
    ....ok?
     
  16. Sep 2, 2010 #15
    The easiest way to look at it is as the set of all vectors that are orthogonal to V=(-1, -2, 1):

    (-1,-2,1) . (a, b, c) = c - a - 2b = 0

    Then if V.x = 0 and V.y = 0 then V.(x+y) = V.x + V.y = 0 etc... so it's a vector space.
     
  17. Sep 2, 2010 #16
    Take two arbitrary vectors of the form (a,b,c) such that the 3rd component of each of these vectors minus its 1st component equals two times its 2nd component. You could call them (a, b, c) and (d, e, f), where each of these letter can be any real number. So c - a =2b, and f - d = 2e. Add them together in the usual way: (a, b, c) + (d, e, f) = (a + d, b + e, c + f). Check whether the 3rd component of (a + d, b + e, c + f) minus its 1st component is equal to two times its 2nd component. If so, then it belongs to your set. The set would then be "closed under addition", one of the requirements of a vector space. Simon_Tyler showed a neat way of doing this test.

    You also need your set to be closed under scalar multiplication, that is, when you multiply a vector in the set by any real number, the result must also be in the set. Take an arbitrary vector (a, b, c), such that c - a =2b, and multiply it by an arbitrary real number: s (a, b, c) = (sa, sb, sc), then ask yourself whether sc - sa = 2sb, given that c - a =2b. If so, its closed under scalar multiplication.

    Finally, check whether the zero vector (0, 0, 0) belongs to your set, that is, does the 3rd component of (0, 0, 0) minus its 1st component equals two times its 2nd component? All vector spaces must have a zero vector. If so, you've got yourself a vector space! (Actually, in this example, this condition is already covered by either of the first two tests. Can you see why?)
     
    Last edited: Sep 2, 2010
  18. Sep 3, 2010 #17

    HallsofIvy

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    You are still wording that badly. What you have written implies that you have some arbitrary <a, b, c> and want to show that c- a= 2b. Of course, you cannot do that- it's not true. You have your "hypotenuse" and "conclusion" reversed.

    What you want to prove is that if a, b, and c are such that c- a= 2b, then the set of all such <a, b, c> form a vector space, with the usual vector addition and scalar multiplication- that is, it is a subspace of [itex]R^n[/itex].

    From c- a= 2b, we have c= a+ 2b so that <a, b, c>= <a, b, a+ 2b>. Another vector of that form is <x, y, x+ 2y>. The sum of any two such vectors is <a, b, a+ 2b>+ <x, y, x+ 2y>= <a+ x, b+ y, a+ 2b+ x+ 2y>= <a+ x, b+ y, (a+ x)+ 2(b+ y)> which is again of the same form. Also, if r is any real number, then r<a, b, a+ 2b>= <ra, rb, r(a+ 2b)>= <ra, rb, ra+ 2rb>, again in that subspace.
     
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