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Polar Cones basic property

  1. Nov 1, 2012 #1
    Let S1*(S2*) be the polar cone of the set S1(S2) (http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone).

    How can I show that if S1 is contained in S2 then S2* is contained in S1*.

    It looks obvious (especially if we think in R^2), but I do not find a way to prove it.
     
  2. jcsd
  3. Nov 1, 2012 #2

    tiny-tim

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    welcome to pf!

    hi avilaca! welcome to pf! :smile:
    isn't the proof obvious from the definition based on inner product? (see http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone) :wink:
     
  4. Nov 3, 2012 #3
    Ok, it's easy from the inner product <a,s> = ||a||.||s||cos[itex]\theta[/itex].
    <a,s> [itex]\leq[/itex] 0 <=> pi/2 [itex]\leq[/itex] [itex]\theta[/itex] [itex]\leq[/itex] 3pi/2.
    This means that if S1 [itex]\subset[/itex] S2, by the above result, the region where the condition {<a,s> [itex]\leq[/itex] 0 , s [itex]\in[/itex] S1 or S2, a [itex]\in[/itex] ℝ[itex]^{n}[/itex]} is true for S1 is the same or it's larger than the one for S2, wich implies S2* [itex]\subset[/itex] S1*.
     
  5. Nov 3, 2012 #4
    Now another challenge:

    Let S = {x [itex]\in[/itex] ℝ[itex]^{n}[/itex]: x = Ap, p [itex]\geq[/itex] 0}, where A [itex]\in[/itex] M[itex]_{n*m}[/itex], p [itex]\in[/itex] ℝ[itex]^{m}[/itex].
    What is its polar cone S*?
     
  6. Nov 3, 2012 #5

    tiny-tim

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    looks good! :smile:
    show us what you get :wink:
     
  7. Nov 4, 2012 #6
    I didn't achieve a great conclusion.

    S* can be defined by {a [itex]\in[/itex] R[itex]^{n}[/itex]: x[itex]^{T}[/itex]a [itex]\leq[/itex] 0, for all x [itex]\in[/itex] S}.

    Now:
    x[itex]^{T}[/itex]a [itex]\leq[/itex] 0 <=> (Ap)[itex]^{T}[/itex]a [itex]\leq[/itex] 0 <=> p[itex]^{T}[/itex]A[itex]^{T}[/itex]a [itex]\leq[/itex] 0.

    But I can not conclude nothing about "a" from here.
     
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