1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector-related Proof

  1. Feb 10, 2007 #1
    Hi all,

    While working on my differential equations homework, I encountered a proof dealing with linear independence and vector addition. I sort of know how to proceed but, not having dealt with formal proofs much, I am afraid that I may not be addressing all necessary apects of the proof. Anyway, here is the question: Prove that if the vectors x = (x_1)i + (x_2)j and y = (y_1)i + (y_2)j
    are linearly independent, then any vector z = (z_1)i + (z_2)j can be expressed as a linear combination of x and y.

    The linear combination of x and y gives us (x_1)i + (x_2)j + (y_1)i + (y_2)j. Rearranging terms, [(x_1)+(y_1)]i + [(x_2)+(y_2)]j = x+y. We can now define x+y = z. Therefore, z = (z_1)i + (z_2)j

    Where do I bring in the necessity of linear independence?
     
  2. jcsd
  3. Feb 10, 2007 #2

    radou

    User Avatar
    Homework Helper

    Actually, this question is kind of 'definition-like'. You have two vectors in V^2(O). Any set of two vectors in V^2(O) which are linearly independent (i.e. non collinear) form a basis for V^2(O), and hence every vector from V^2(O) can be represented uniquely as a linear combination of these two independent vectors.
     
  4. Feb 10, 2007 #3
    What exactly is V^2(O)? I haven't had linear algebra so I am probably unfamiliar with some terminology.

    EDIT: Is that just a 2-D vector space?
     
    Last edited: Feb 10, 2007
  5. Feb 10, 2007 #4

    radou

    User Avatar
    Homework Helper

    My apologies for not pointing it out - V^2(O) (or call it whatever you like) is the set of all radius vectors in the Euclidean plane, where your story is, of course, set up in a Cartesian coordinate system.

    EDIT: it could be, but be careful when using that terminology; formally, a 2-D vector space is any 2-dimensional vector space - its elements don't need to be radius vectors!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Vector-related Proof
  1. Relation proof (Replies: 1)

  2. Vector Proof (Replies: 0)

  3. Vector Proof (Replies: 2)

  4. Vector Proof (Replies: 11)

  5. Relations proof (Replies: 2)

Loading...