Can anybody check my answers (linear algebra)?

[Artusartos]

Homework Statement

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.

Homework Equations

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

 

[mighty2000]

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Hi, your answers seem to be mostly correct, but there are a few things that could be clarified.

1) In order to show that Ann_A(V) \subset Ann_A(v), you need to show that for every g in Ann_A(V), g must also be in Ann_A(v). Your reasoning is correct, but you should make it more explicit. You can say something like "Let g be an element of Ann_A(V). This means that g(A)w = 0 for all w in V. Since v is also in V, we have g(A)v = 0. Therefore, g is also in Ann_A(v), so Ann_A(V) \subset Ann_A(v)."

2) Your reasoning for the first part is correct, but again, you should make it more explicit. You can say something like "Let g be an element of [g \in F[x]: g(T) = 0]. This means that g(T)w = 0 for all w in V. But since T(w) = Aw for all w in V, we have g(T)w = g(A)w = 0 for all w in V. Therefore, g is also in [g \in F[x]: g(A)w = 0 for all w \in V], so [g \in F[x]: g(T) = 0] \subset [g \in F[x]: g(A)w = 0 for all w \in V]."

For the second part, you should mention that [T]_B is the matrix representation of T with respect to the basis B, and that [T]_B is in V because T is a linear transformation on V. Also, you should mention that g([T]_B) is the matrix representation of g(T) with respect to the basis B. So the final result should be "Therefore, [g \in F[x]: g(T) = 0] = [g \in F[x]: g([T]_B) = 0]."