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Little excercise about bases (question)

  1. May 17, 2005 #1
    Suppose u1, u2 and u3 are a basis for R^3.

    So there is a unique combination
    c1.u1 + c2.u2 + c3.u3 = v

    for every v in R^3.

    Now suppose I have this bunch of vectors (u1+u2), (u1+u3), (u2+u3) and I would like to know if they also are a basis for R^3.
    So v = d1.(u1+u2) + d2.(u1+u3) + d3.(u2+u3)
    and (d1,d2,d3) has to be unique, right?

    This will come out nicely in this example, I would get 3 different and unique solutions for d1, d2 and d3.

    Now my question actually is: is it possible (in some other excercise) to have a unique solution for (d1,d2,d3), so that my new bunch of vectors turns out to be a basis, but that d2 has the same value as d3? And what does this actually mean?

    I've been thinking about this for a while, but can't seem to figure it out. :confused:
  2. jcsd
  3. May 17, 2005 #2


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    That is a part of it.A basis is complete (which is what u suggested),but the vectors need to be linear independent as well.

    Do u know the condition for linear independence...?(You'd better,else it's useless to know what u've written above).

  4. May 18, 2005 #3
    Okay, I know that when u1,..,un is a basis, every vector v can be written as a unique combination of this basis. Never really thought about the opposite, but that's what I assumed in this excercise because that's what I learned at school.

    So I want to prove that:
    If the combination for v = c1.u1 +..+ cn.un is unique => u1,..,un is a basis.

    So I know that if u1,..,un is linear independent AND generates every v, it is a basis.

    1) I know that it generates v, because that's what the combination tells me.
    2) Now I would like to express 0 as a combination of u1,..,un. One way to do this is 0.u1 +..+ 0.un = 0. But wait, I start with the fact that every combination is unique! So if I want to look for 0, my (c1,..,cn) = (0,..,0) so u1,..,un is linear independent.

    Can anyone confirm that I did this the right way? I hope I did. :smile:

    I also found this website: http://www.answers.com/topic/basis-linear-algebra
    Seems that condition 1 and 4 are equivalent.
    Last edited: May 18, 2005
  5. May 18, 2005 #4


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    Heh,you didn't read very carefully.All 4 of them are equivalent.It says so right there.

  6. May 18, 2005 #5
    I used condition 4 to solve my excercise, and thought you were talking about condition 1, hence my reply.
    But maybe I misinterpreted? If so, my apologies.

    Still, I don't know if it's possible that d2 = d3. If you have the answer, you can always PM it to me in Dutch if you want..maybe that's easier.
  7. May 18, 2005 #6


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    Nope,all Fourier coeff.have to be unique,not necessarily different.Think of a vector in [itex] \mathbb{R}^{3} [/itex] along Ox.What are the F.coeff.associated to the basis vectors [itex] \vec{j}[/itex] and [itex] \vec{k}[/itex] ?

  8. May 18, 2005 #7

    But then I'm not talking about every v = c1.u1 +..+ cn.un to begin with? I'm only talking about this line of vectors. So I start off with a restriction on the original coefficients?

    So in R3, if d2=d3 => I can't describe all the vectors.
    Can I use contraposition on this?
    If I can describe all vectors => solutions d have to be different

    I have a feeling this is probably horse poo. :biggrin:
  9. May 18, 2005 #8


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    I don't get it."So in R3, if d2=d3 => I can't describe all the vectors" doesn't make any sense.

  10. May 18, 2005 #9
    In your example vectors have coordinates (x,0,0). So in any basis of R3 they would together represent a line.

    If I start with a basis in R3: u1,..,un then v = c1.u1 +..+ cn.un represents any vector in R3.

    How could I ever find that my new coordinates are (x,0,0)? That would mean that some of those vectors v can't be described with the new basis.

    If I have (d1,d2,d2) then I would have a plane, and some of the vectors couldn't be described also.

    Does that make sense?
  11. May 18, 2005 #10


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    That would be a rotation of coordinate axis.The line would be the same,just the basis in [itex] \mathbb{R}^{3} [/itex] would be another.

  12. May 18, 2005 #11
    Yes, of course. But I'm talking about the initial question in the first post here.

    Suppose v is ANY vector in R3, so not just that line. And suppose I rotate the coordinate axis. Then v has a unique combination in this new basis, for example: v = d1.e1 + d2.e2 + d3.e3
    Now what I'm saying is that, IF d1 = d2 then co(v)=(d1,d2,d2) in the new basis and this means that I'm dealing with a plane instead of the whole 3D-space, correct?

    Since that is a contradiction (v is any vector in R3, not just a plane of vectors), it's necessary that every d is unique AND different (reductio ad absurdum).

    That's what I'm wondering about. Is my reasoning correct, or flawed?
  13. May 18, 2005 #12


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    What is that...?co(d1,d2,d3)...?:bugeye:

  14. May 18, 2005 #13
    co(v) = the coordinate of v in basis e1,e2,e3 :shy:
  15. May 18, 2005 #14


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    Why plane...?You're mixing elements of geometry with the alegbra ones.

    [tex] \vec{v}\in\mathbb{R}^{3} [/tex] (1)

    [tex] \vec{v}=v^{i}\vec{e}_{i} [/tex] (2)

    [tex] \vec{v}=v'^{i}\vec{e'}_{i} [/tex] (3)

    [tex] \vec{e}_{i}=A_{i}{}^{j} \vec{e'}_{j} [/tex] (4)

    ,where [tex] \hat{A} [/itex] is an orthogonal 3*3 matrix,[tex] \hat{A}\in\mbox{O(3)} [/tex].

    Last edited: May 18, 2005
  16. May 18, 2005 #15
    Okay, clearly I haven't got a clue, because now I don't know what you're talking about. :frown:

    I don't quite understand the notations in 2, 3 and 4. I guess A is the rotation matrix, but I don't see how this contradicts what I suggested.
  17. May 18, 2005 #16


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    It doesn't.It's a mathematical form of the point i'm trying to make.

    The vectors in the basis are still independent,it doesn't really matter what the values of the components are.I showed you an example where 2 of the 3 comp are zero.You can cook a vector in the Oxy plane and its "z" comp.will be zero,too.

    I'm not really seeing your point.

  18. May 18, 2005 #17
    Okay, let me show you how I got to the question in the first place. For that I need to work out the excercise:

    given: u1,u2,u3 is a basis of R3
    to prove: (u1+u2),(u1+u3),(u2+u3) is a basis

    Suppose v = c1.u1 + c2.u2 + c3.u3 for every v in R3 (with c's are unique)

    also: v = d1.(u1+u2) + d2.(u1+u3) + d3.(u2+u3) (d's unique)
    <=> v = (d1+d2).u1 + (d1+d3).u2 + (d2+d3).u3
    /d1+d2 = c1
    |d1+d3 = c2
    \d2+d3 = c3

    and I find:

    /d1 = (-c1 + 3.c2 + c3)/2
    |d2 = (c1 - c2 + c3)/2
    \d3 = (-c1 + c2 + c3)2

    So I find these unique solutions for d and everything is great!

    I started wondering, what if d2 = d3 = (c1 - c2 + c3)/2 ? "Does there exist some excercise of the same sort that gives me the same solution for d2 and d3", I asked myself. It doesn't really have to be this particular one, but any value for d2 (for example (c1-c2)/2 or something).

    Now I come to think of it, this doesn't mean I'm rotating the basis at all does it.. I'm just using another one, any one.. So whatever I said above about rotation, forget it. :cool:

    So, instead of saying v has coordinates (d1,d2,d3) in this new basis, I can say v has coordinates (d1,d2,d2). These vectors v form a plane. Why do I say plane here? If I go d2 times in the direction of the second basis vector, and also d2 times in the direction of the third basis vector for any value d2, I would only get solutions on a straight line (like the 45° line if I'm working in the standard basis).
    Now if I add any vector on this line to d1 times the first basis vector, I get a plane in 3D. So this combination only describes the vectors that lie in this plane and not the whole R3 as was supposed to.

    Of course I'm going to get some vectors that have two of the same coordinate numbers, because I'm in the whole R3, but that's a specific case and what I'm talking about is that the vectors have two of the same coordinate numbers in the general case.

    Are you able to see what I mean now?
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