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Dimensionality, Rangespace & Nullspace Problem

  1. May 7, 2008 #1
    [SOLVED] Dimensionality, Rangespace & Nullspace Problem

    The problem statement, all variables and given/known data
    Prove (where A is an n x n matrix and so defines a transformation of any n-dimensional space V with respect to B, B where B is a basis of V) that [itex]\dim(R(A) \cap N(A)) = \dim R(A) - \dim R(A^2)[/itex]

    The attempt at a solution
    If I determine the basis of [itex]R(A) \cap N(A)[/itex], I can determine its dimensionality and then compare it with [itex]\dim R(A) - \dim R(A^2)[/itex].

    I've been unsuccessful at finding a basis. Also, given that [itex]\dim R(A) = m[/itex], is there a way to determine what [itex]\dim R(A^2)[/itex] is?
     
  2. jcsd
  3. May 8, 2008 #2
    Let [tex] \{Av_1, \ Av_2, \ ..., \ Av_k\} [/tex] be a basis for [tex]R(A)\cap N(A)[/tex] and [tex] \{A^2u_1, \ A^2u_2, \ ..., \ A^2u_m\} [/tex] be a basis for [tex]R(A^2)[/tex], with [tex] v_1, \ v_2, \ ..., \ v_k, \ u_1, \ u_2, \ ..., \ u_m \in V[/tex].

    We claim that X = [tex]\{Av_1, \ Av_2, \ ..., \ Av_k, \ Au_1, \ Au_2, \ ..., \ Au_m\}[/tex] is a basis for [tex]R(A)[/tex].

    Proof:
    1) X spans [tex]R(A)[/tex].

    Take any [tex] w \in R(A)[/tex]. Since [tex]Aw \in R(A^2)[/tex], so [tex]Aw = b_1 A^2u_1 \ + b_2 A^2u_2 \ + ... \ + b_m A^2u_m[/tex], for some scalars [tex]b_1, \ b_2, \ ..., \ b_m[/tex].

    Thus, [tex]A(w \ - b_1 Au_1 \ - b_2 Au_2 \ - ... \ - b_m Au_m) = \mathbf{0}[/tex] and so [tex] w \ - \ b_1 Au_1 \ - \ b_2 Au_2 \ - \ ... \ - \ b_m Au_m \in R(A) \cap N(A)[/tex] (Why?)
    Complete the proof yourself :)

    Hence, every element in [tex]R(A)[/tex] can be expressed as a linear combination of [tex]Av_1, \ Av_2, \ ..., \ Av_k, \ Au_1, \ Au_2, \ ..., \ Au_m[/tex] and so X spans [tex]R(A)[/tex].

    2) X is a linearly independent set.
    We solve the homogeneous equation [tex]c_1 Av_1 \ + \ c_2 Av_2 \ + \ ... \ + \ c_k Av_k \ + \ c_{k+1} Au_1 \ + \ c_{k+2} Au_2 \ + \ ... \ + \ c_{k+m} Au_m = \mathbf{0}[/tex], where [tex]c_1, \ c_2, \ ..., \ c_{k+m}[/tex] are scalars.

    Now, [tex]A(c_1 Av_1 \ + \ c_2 Av_2 \ + \ ... \ + \ c_k Av_k \ + \ c_{k+1} Au_1 \ + \ c_{k+2} Au_2 \ + \ ... \ + \ c_{k+m} Au_m) = \mathbf{0}[/tex], and thus...
    Complete the proof yourself: I suggest first showing that [tex]c_{k+1} \ = c_{k+2} \ = ... \ = c_{k+m} \ = 0[/tex]

    Since the homogeneous equation has only the trivial solution, X is a linearly independent set.

    This proves our claim and the result follows.
     
  4. May 8, 2008 #3
    Neat. I wish I had your intuition.

    w is in R(A) and so is [tex]- b_1 Au_1 - \cdots - b_m Au_m[/tex] because it's a linear combination of members of X. Thus [tex]w - b_1 Au_1 - \cdots - b_m Au_m[/tex] is in R(A). Since [tex]A(w - b_1 Au_1 - \cdots - b_m Au_m) = \mathbf{0}[/tex], [tex]w - b_1 Au_1 - \cdots - b_m Au_m[/tex] is also in N(A).

    Let [tex]\mathfrak{v} = c_1 Av_1 + \cdots + c_k Av_k[/tex] and let [tex]\mathfrak{u} = c_{k+1} Au_1 + \cdots + c_{k+m} Au_m[/tex]. [tex]A(\mathfrak{v} + \mathfrak{u}) = A\mathfrak{v} + A\mathfrak{u} = A\mathfrak{u} = \mathbf{0}[/tex].

    The last equality above plus the fact that [tex]\{A^2u_1, \ldots, A^2u_m\}[/tex] is linearly independent means that [tex]c_{k+1} = \cdots = c_{k+m} = 0[/tex]. Thus [tex]\mathfrak{v} + \mathfrak{u} = \mathbf{0}[/tex] reduces to just [tex]\mathfrak{v} = \mathbf{0}[/tex] and since [tex]\{Av_1, \ldots, Av_k\}[/tex] is a linearly independent set, [tex]c_1 = \cdots = c_k = 0[/tex] as well. Ergo, X is linearly independent.
     
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