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Linear algebra Homework Problem

  1. Jul 8, 2007 #1
    There doesn't seem to be a homework problem section for linear algebra so either this is the place to post it or they'll move it.

    The question is to prove that r0 = 0. I can't even figure out the first step. Can someone tell me the first step in solving this problem? Thanks.

    Pete
     
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Jul 8, 2007 #2

    jambaugh

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    You should start with the property defining the zero vector or matrix:
    0 + X = X

    Can you use the assumed properties of scalar multiplication and vector/matrix addition to show this is true for r0?
     
  4. Jul 8, 2007 #3
    That's what I was trying to do with no success.

    Pete
     
  5. Jul 8, 2007 #4

    radou

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    Let a be a scalar, and 0 the zero vector in some vector space.

    a0 + a0 = ...
     
  6. Jul 8, 2007 #5

    HallsofIvy

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    Are you aware that r(u+ v)= ru+ rv?

    Use that with u= 0 and it's easy.

    By the way, there is a place for homework like this- it is the "Calculus and Beyond" group. I'm moving it.
     
  7. Jul 9, 2007 #6
    I'm aware of all the axioms of linear algebra (I hope). :biggrin: After all I had to take this course in College in order to graduate with a BA in Math (along with a BA in physics - double major)
    I don't see how its easy since it goes from

    ru + rv = ru + rv

    Setting u = 0 to obtain

    (u + v) = rv

    There's not even a zero vector in there to work with. Can you work out your idea here?

    I also think its a bad idea to not allow homework to be placed here in the future. The people who can help on homework the best are those people who post all the time in their group of expertise. A place for homework would not draw attention to all types of homework questions. The question could be stated here and then taken offline (to PM) to discuss it. That sounds like the better idea, in my humble opinion that is.

    Best regards

    Pete
     
    Last edited: Jul 9, 2007
  8. Jul 9, 2007 #7

    matt grime

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    Calculus and Beyond means US College level questions. (Whether or not you like that distinction.) So post there.

    Secondly, how hard did you play around with the suggestions? Let's give you an even bigger hint: you're allowed to let u=v.

    0.(u+u)=0.(2.u)=0*2.(u)=0.u, right? Now what about the other way to write 0.(u+u)?
     
  9. Jul 9, 2007 #8
    Hello Pete,

    may I ask if you mean with 0 the vector (0,0,......0)?
    And is r an element of [itex]\mathbb{R}[/itex]?
     
  10. Jul 9, 2007 #9

    HallsofIvy

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    Are you joking? I said "set u= 0". Don't do it on just one side of the equation!
    r(v+ 0)= rv+ r0. Now, since v+0= v, what is r(v+0)?

     
  11. Jul 9, 2007 #10

    HallsofIvy

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    I don't believe he was assuming that- the statement is true and easy to prove in any vector space. r is an element of what ever field the vector space is based on.
     
  12. Jul 9, 2007 #11

    jambaugh

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    Try: r0+X = r(0 + 1/rX) = ...
     
  13. Jul 10, 2007 #12
    Hi Edgardo

    Yes 0 = (0,0,......0) (number of zeros = dimension of the space you're working in).

    Yes r an element of [itex]\mathbb{R}[/itex]?

    Thanks for all you help folks but today I have a lot of things that have to get done so I can't work on this problem today. I'll be back on this either tonight or tomorrow morning etc.

    Pete
     
    Last edited: Jul 11, 2007
  14. Jul 11, 2007 #13
    what about proof by definition?
    the length of vector 0 is zero, hence the length of vector 0r is also 0 => vector 0r is 0
     
  15. Jul 11, 2007 #14

    HallsofIvy

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    I was under the impression that this was a problem in Linear Algebra. It would make no sense at all to restrict to Rn or even to "normed" spaces (in which "length" is defined).

    huyen vyvy, even in a normed space, do you consider it easier to prove "r times a vector of 0 length has 0 length" than "r times the 0 vector is the 0 vector"? Seem much harder to me.


    In any case, as I pointed out long ago (long enough that I assume any homework is already overdue), r(u+ v)= ru+ rv. Let v= 0. Then r(u+ 0)= ru+ r0. Since u+ 0= u (definition of 0), that is ru= ru+ r0. Adding -(ru) to both sides (every vector has an additive inverse) leaves 0= r0.

    That's about as easy as it gets.
     
  16. Jul 12, 2007 #15
    Hi Pete,

    the first thing that came into my mind was to use the definition
    for vectors in R^n:

    r 0 := (r0, r0, ....,r0) = (0, 0, ....0) = 0

    But as HallsofIvy already mentioned, r 0 = 0
    for every vector space.

    HallsofIvy gave a proof for this above.
    Similarly, you can use the fact that 0 + 0 = 0.
     
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