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Matrices on Linear Subspaces

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Let's say I'm given two vectors
    [tex] v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \in \mathbb R^4[/tex].
    Let W be subspace spanned by these vectors, and define [itex] G = v_1 v_1^T + v_2 v_2^T [/itex] a matrix mapping [itex]\mathbb R^4 \to \mathbb R^4 [/itex]. Find [itex] G' [/itex] such that [itex] \left. G' = G^{-1} \right|_{W} [/itex].


    2. Relevant equations



    3. The attempt at a solution

    Since [itex] v_1, v_2 [/itex] are linearly independent, the dimension of W is 2. Furthermore, since G is composed of these vectors, we can be guaranteed that that an inverse exists on W. Let W' be the image of W under G. That is, since [itex] v_1, v_2 [/itex] generate W, then [itex] G v_1, Gv_2 [/itex] should generate W'.

    I've computed that
    [tex] G = \begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, Gv_1 = \begin{pmatrix} 2\\ 3\\ 1 \\0 \end{pmatrix}, Gv_2 = \begin{pmatrix} 1 \\ 3 \\ 2 \\ 0 \end{pmatrix} [/tex].

    Now maybe it's because it's been so long since I did any linear algebra, but I can't for the life of me figure out how to "extract" the restriction of G to W. Given this information, it should then be simple to construct the inverse and hence make G'.
     
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  3. Sep 29, 2009 #2

    Dick

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    Just express Gv1 and Gv2 as linear combinations of v1 and v2. At some point you should figure out you really are doing this the long way around. Gv1=v1(v1)^Tv1+v2(v2)^Tv1. I see dot products in there.
     
  4. Sep 30, 2009 #3
    So what I think you're saying is that I have

    [tex] v_1' = G v_1 = 2 v_1 + v_2 [/tex]
    [tex] v_2' = G v_2 = v_1 + 2v_2 [/tex]

    So I should associate [itex] v_1' = (2,1), v_2' = (1,2) [/itex]. In this case then, if G is determined by its action on the basis states [itex] v_1, v_2 [/itex] then G acting on W can be written as

    [tex] G_W = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} [/tex]

    Which in turn has inverse

    [tex] G_W^{-1} = \frac13 \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} [/tex]

    But then how do I make this four-dimensional again?
     
  5. Sep 30, 2009 #4

    Dick

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    You can make it four dimensional any way you want. You know how (G_W)^(-1) acts on v1 and v2. You'll have to pick v3 and v4 so that {v1,v2,v3,v4} is a basis for R^4. Now you can define (G_W)^(-1)(v3) and (G_W)^(-1)(v4) to be anything you want. You know you can't extend it to be G^(-1) on R^4, since G isn't invertible, right?
     
  6. Sep 30, 2009 #5
    Is there no loss of information or structure defined by the original G by choosing v3 and v4 as such?
     
  7. Sep 30, 2009 #6
    Okay, let me see if I understand this then:

    [tex] G_W^{-1} v_1 = \frac13 \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 2 v_1 -v_2 [/tex]
    [tex] G_W^{-1} v_2 = \frac13 \begin{pmatrix} -1 \\ 2 \end{pmatrix} = -v_1 + 2 v_2 [/tex]

    Thus we use our original definitions of
    [tex] v_1 = \begin{pmatrix} 1 \\ 1\\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} [/tex]
    To find that the first two columns of G' should be

    [tex] \frac13 \begin{pmatrix} 2 & - 1 \\ 1 & 1 \\ -1 & 2 \\ 0 & 0 \end{pmatrix} [/tex]

    But now you say that [itex] \{ v_1, v_2, v_3, v_4 \} [/itex] must form a basis for [itex] \mathbb R^4 [/itex], but that I can choose [itex] G_W^{-1} v_3 , G_W^{-1} v_4 [/itex] to be anything I want. This seems contradictory, since I could certainly choose these such that [itex] v_3, v_4 [/itex] did not help form a basis. So should I choose them such that they form a basis? Or choose them arbitrarily?
     
  8. Sep 30, 2009 #7

    Dick

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    I'm not really sure what you are trying to do. [itex]\left. G' = G^{-1} \right|_{W}[/itex], doesn't make much sense because [itex]G^{-1}[/itex] doesn't exist. G maps R^4->W. [itex]{G \right|_W}^{-1}[/itex] exists, but it doesn't have any unique extension to R^4.
     
  9. Sep 30, 2009 #8
    Sorry, I wrote that backwards. It should be

    [tex] \left. G_W^{-1} = G' \right|_{W} [/tex]
     
  10. Sep 30, 2009 #9

    Dick

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    I still don't see how that would be unique if you want G' to be 4x4.
     
  11. Sep 30, 2009 #10
    Perhaps some perspective will help. I have a series of density matrices [itex] \{ \rho_i \} [/itex] and my goal is to construct a positive operator valued measure from these states to form a measure on [itex] (\mathbb C^3)^{\otimes 3} = \mathbb C^9 [/itex]. This is done by defining [itex] M = \sum_i p_i \rho_i [/itex] where [itex] p_i [/itex] are the associated density matrix probabilities. Then the positive operator valued measure is defined as

    [tex] E_i = p_i M^{-\frac12} \rho_i M^{-\frac12} [/tex]

    where [itex] \left( M^{-\frac12} \right)^2 M [/itex] is the projection operator onto the image of M.
     
  12. Sep 30, 2009 #11

    Dick

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    Sorry, I really don't know that formalism.
     
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