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Stuck on this Algebra Problem

  1. Nov 1, 2004 #1

    AKG

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    T be a linear operator (I think they mean "Let T be a linear...") on a finite-dimensional vector space V, and let [itex]W_1[/itex] be a T-invariant subspace of V. Let [itex]x \in V[/itex] such that [itex]x \notin W_1[/itex]. Prove the following results:
    • There exists a unique monic polynomial [itex]g_1(t)[/itex] of least positive degree such that [itex]g_1(T)(x) \in W_1[/itex].
    • If h(t) is a polynomial for which [itex]h(T)(x) \in W_1[/itex], then [itex]g_1(t)[/itex] divides h(t).
    • [itex]g_1(t)[/itex] divides the minimal and characteristic polynomials of T.
    • Let [itex]W_2[/itex] be a T-invariant subspace of V such that [itex]W_2 \subseteq W_1[/itex], and [itex]g_2(t)[/itex] be the unique monic polynomial of least degree such that [itex]g_2(T)(x) \in W_2[/itex]. Then [itex]g_2(t)[/itex] divides [itex] g_2(t)[/itex].
    • If [itex]\beta[/itex] is a basis of V, and [itex]\beta _{W_1}[/itex] is a basis of [itex]W_1[/itex] such that [itex]\beta _{W_1} \subseteq \beta[/itex], then define [itex]W_1 \prime = Span(\beta - \beta _{W_1})[/itex]. [itex]V = W_1 \oplus W_1 \prime[/itex]. [itex]\forall v \in V, \exists w_1 \in W_1, w_1 \prime \in W_1 \prime[/itex] such that [itex]v = w_1 + w_1 \prime[/itex].

      [tex]g_1(T)(x) = g_1(T)(w_1 + w_1 \prime) = g_1(T)(w_1) + g_1(T)(w_1 \prime)[/tex]

      Now, if the restriction of T to [itex]W_1 \prime[/itex] were an operator on [itex]W_1 \prime[/itex], then there would be a unique monic polynomial of least degree such that [itex]g_1(T)(w_1 \prime) = 0[/itex], namely the minimal polynomial of T restricted to [itex]W_1 \prime[/itex]. Then, if [itex]g_1(t)[/itex] is a polynomial over the same field that underlies [itex]W_1[/itex], I can assert that [itex]g_1(T)(w_1) \in W_1[/itex], and thus the result is proved. Can I prove these "if"s? If not, is there another way to prove the result? I haven't looked at the rest of it yet.
     
    Last edited: Nov 1, 2004
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  3. Nov 1, 2004 #2

    AKG

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    Well, assuming that it's true that [itex]W_1'[/itex] is T-invariant, more can be proven:

    (b) If [itex]h(T)(x) \in W_1[/itex], [itex]h(T)(w_1') = 0\ \forall w_1' \in W_1'[/itex], so if [itex]T_{W_1'}[/itex] is the restriction of T to [itex]W_1'[/itex], then [itex]h(T_{W_1'}) = T_0[/itex]. My book proves that if [itex]g_1(t)[/itex] is the minimal polynomial of a linear operator [itex]T_{W_1'}[/itex] on a finite-dimensional vector space [itex]W_1'[/itex], then for any polynomial h(t), if [itex]h(T_{W_1'}) = T_0[/itex], [itex]g_1(t)[/itex] divides [itex]h(t)[/itex]. Of course, insted of "[itex]W_1'[/itex]", my book proves it for "V", and instead of the minimal polynomial being "[itex]g_1(t)[/itex]" it calls it "p(t)" in the proof, but the result still holds.

    (c) It suffices to show that [itex]g_1(t)[/itex] divides the minimal polynomial of T, since the minimal polynomial of T already divides the characteristic polynomial of T (a corollary of the theorem from my book that I just mentioned). Let p(t) be the minimal polynomial of T. Then,

    [tex]\forall v \in V,\ v = w_1 + w_1',\ w_1 \in W_1,\ w_1' \in W_1':[/tex]

    [tex]p(T)(v) = p(T)(w_1) + p(T)(w_1') = 0[/tex]

    Since [itex]W_1[/itex] and (supposedly) [itex]W_1'[/itex] are T-invariant, and since (supposedly) p(t) takes coefficients from the same field that underlies [itex]W_1[/itex] and [itex]W_1'[/itex], we have the above equality holding if and only if:

    [tex]p(T)(w_1) = 0,\ p(T)(w_1') = 0[/tex]

    Since this must be true for all [itex]w_1'[/itex], [itex]p(T_{W_1'}) = T_0[/itex], so, from (b), [itex]g_1(t) | p(t)[/itex].

    (d) Define [itex]W_2'[/itex] in a way similar to how [itex]W_1'[/itex] was defined. Let [itex]g_2(t)[/itex] be the minimal polynomial of T restricted to [itex]W_2'[/itex]. Then it is the unique polynomial which satisfies the hypotheses, and since [itex]W_2 \subseteq W_1[/itex], [itex]g_1(t) | g_2(t)[/itex] by (b).

    All of this is pretty simple, but it all rests on two (huge) assumptions, mainly the assumption that if [itex]W_1[/itex] is T-invariant, then so is [itex]W_1'[/itex].
     
    Last edited: Nov 1, 2004
  4. Nov 1, 2004 #3

    AKG

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    I think I'm on to something I can use, but too tired to figure out exactly what to do with it. Anyways, define S to be the projection of V on [itex]W_1[/itex], and U to be the projection of V on [itex]W_1 \prime[/itex]. If we define [itex]g_1(t)[/itex] to be the minimal polynomial of UT, I think we'll have something.

    T = ST + UT
    [itex]g_1(T) = g_1(ST + UT)[/itex]
    [itex]g_1(T) = a_n(ST + UT)^n + \dots a_0I[/itex]

    Now if any power of ST is composed with UT, we will get zero since SU = 0, so:

    [itex]g_1(T)= g_1(ST) + g_1(UT) = g_1(ST)[/itex].

    Now UT really is an operator on the vector space [itex]W_1 \prime[/itex], that is, [itex]W_1 \prime[/itex] really is UT-invariant. I believe the rest should hold with a few adjustments. Now, is it problematic that I assume that p(t) takes coefficients from the same field as the one that underlies the subspaces?
     
  5. Nov 1, 2004 #4

    shmoe

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    No you can't. V any vector space, [tex]W_1[/tex] any non-trivial subspace, T the projection map on [tex]W_1[/tex]. You can't pick a [tex]W_2[/tex] in the manner you describe to be T-invariant.

    Think quotient space.
     
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