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Homework Help: Can anybody check my answers (linear algebra)?

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]A \in M_n(F) [/itex] and [itex]v \in F^n [/itex].

    Also...[itex][g \in F[x] : g(A)(v)=0] = Ann_A (v) [/itex] is an ideal in F[x], called the annihilator of v with respect to A. We know that [itex] g \in Ann_A(v) [/itex] if and only if f divides g in F[x]. f is the monic polynomial of lowest degree in the set...so it is the minimal polynomial and divides everything in the set. Let [itex]V = Span(v, Av, A^2v, ... , A^{k-1}v).[/itex]. V is the smallest A-invariant subspace containing v. We denote the fact by writing V=F[x]v. This corresponds to the F[x]-module structure on [itex]F^n[/itex] induced by multiplication by A. We also know that [itex]v, Av, A^2v, ... , A^{k-1}v[/itex] is a basis, B, of V.

    Now these are the questions...

    1) Define [itex]Ann_A(V) =[g \in F[x] : g(A)(w) = 0 for all w \in V].[/itex] Show that [itex]Ann_A(V)=Ann_A(v)[/itex]

    2) Let T: V -> V be induced by multiplication by A: T(w)=Aw for [itex]w \in V[/itex]. Show that [itex]Ann_A(V) = [g \in F[x] : g(T) = 0] = [g \in F[x]: g([T]_B)=0][/itex].

    Here the first one means that g(T): V -> V is the 0-transformation and the second one means that g([T]_B) is the 0-matrix. Since [itex]Ann_A(V) = (f) = [g \in F[x]: f|g][/itex], we write [itex]f=min_T(x)[/itex], the monic polynomial of lowest degree with f(T)=0.

    2. Relevant equations

    3. The attempt at a solution

    My answers:

    1) In order to show that [itex]Ann_A(V)=Ann_A(v)[/itex], I need to show that [itex]Ann_A(V) \subset Ann_A(v)[/itex] and [itex]Ann_A(v ) \subset Ann_A(V)[/itex]. It is clear that [itex]Ann_A(V ) \subset Ann_A(v)[/itex].

    In order to show that [itex]Ann_A(V) \subset Ann_A(v)[/itex]...

    we need to show that g.v=0 implies that g.w=0 for all w in V. Since a field is an integral domain, we know that either g or v must be zero. We know that v cannot be zero, because...in the set [itex][g \in F[x] : g(A)(v)=0] = Ann_A (v) [/itex], we know that f is the monic polynomial of lowest degree, and that f dividees every element in that set. If v was equal to zero, then all polynomials with coefficients in F[x] would be in that set...since zero times anything is zero. So we would also have constant polynomials, but f cannot divide a constant polynomial...so that would be a contradiction. So g(A) must be zero. Since g(A) is zero, g(A) times anything is zero...so g(A)w=0 for all w in V.

    2) We say g(T) = g(A), where A is restricted to v.

    [itex]g(x) = c_0 + c_1x +... + c_tx^t [/itex] and [itex] g(T)w= (c_0 + c_1T + ... + c_tT^t)(w) = c_0w + c_1Tw +... + c_t(T)^t)(w) = c_0w + c_1Aw + ... + c_t(A)^tw = (c_0 + c_1A + ... + c_tA^t)(w) = g(A)w [/itex].

    So [itex] [g \in F[x]: g(A).w = 0 for all w \in V][/itex] = [itex][g \in F[x]: g(T).w=0 for all w \in V][/itex] = [itex][g \in F: g(T) = 0][/itex].

    Now for the second one, since we know that [itex][g \in F: g(T) = 0][/itex], we know that g sents every T to zero. Since T is in V, we also know that [itex][T]_B[/itex] is also in V...so g must also sent [itex][T]_B[/itex] to zero.

    Do you think my answers are correct? If not, then can you tell me why?

    Thanks in advance
  2. jcsd
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