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


    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


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook