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Perpendicular Subspace of R^n

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

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    3. The attempt at a solution

    The terminology in this question confuses me into what I am actually trying to solve. It seems to me that S-perp would naturally be a subspace of real column vectors based on the fact that we specify that S[tex]\neq[/tex]0. It goes on to mention that S-perp is nonempty which seems obvious in the fact that S is not empty and it asks to show that any scalar multiples of vectors within the subset of S-perp will continue to be elements of S-perp.

    So i've reached the thought that either
    1) This question is ridiculously simply that intends for me to re-state the obvious or
    2) I've missed something completely and it actually requires a long proof. Any insights?
     
  2. jcsd
  3. Oct 6, 2009 #2

    Mark44

    Staff: Mentor

    Here's an example that might give you some insight. Consider R2, and S = {(x, y) | y = 0}. Sperp = {(x, y) | x = 0}. Each element of Sperp in this example is orthogonal to each element of S, and further, it can be shown that Sperp of this example is a subspace of R2.

    Your problem is similar. You need to show that your Sperp is a subspace of Rn, which means you need to show the following:
    1. Sperp contains the 0 vector.
    2. If u and v are in Sperp, then u + v is also in Sperp.
    3. If u is in Sperp and c is a scalar, then cu is in Sperp.
     
  4. Oct 6, 2009 #3
    Thank you for your assistance. I believe I have constructed a response that is adequate by proving the vectors x,u,v to be elements of S-perp by continually showing that the dot product of the vectors with y (being an element of S) to equal 0 showing orthogonality showing them to be an element of S-perp.
     
  5. Oct 6, 2009 #4

    Mark44

    Staff: Mentor

    That's not going to cut it. All you have shown (based on your description) is that the vectors x, u, and v are in Sperp; you haven't shown that Sperp is a subspace of Rn. To do that, you need to show the three things I listed in my previous post.
     
  6. Oct 7, 2009 #5

    Landau

    User Avatar
    Science Advisor

    Shambles, do you know what a subspace is, or more precisely, how it is defined? That seems to be the problem.
     
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