(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [itex]A \in M_n(F) [/itex] and [itex]v \in F^b [/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|g in F[x]. Let [itex]V = Span(v, Av, A^2v, ... , A^{k-1}v).[/itex]. V is teh 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], since v is in V.

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

we say that, since [itex]Ann_A(V) = [g \in F[x] : g(A)(w) = 0 for all w \in V][/itex], then g(A) must be zero for A restricted to V. So, assuming that it is zero when A is restricted to V, [itex]Ann_A(V) \subset Ann_A(v)[/itex]. So they must be equal.

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, we do the exact same thing, except that we say [itex]g([T]_B) = g(A)[/itex], where g(A) is restricted to [itex][v]_B[/itex].

So...

[itex]g(x) = c_0 + c_1x +... + c_tx^t [/itex] and [itex] g([T]_B)w= (c_0 + c_1[T]_B + ... + c_t{[T]_B}^t)(w) = c_0w + c_1[T]_Bw +... + c_t{[T]_B}^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]_B).w=0 for all w \in V][/itex] = [itex][g \in F: g([T]_B) = 0][/itex]

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

Thanks in advance

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